Question 514 Marks
Solve the following differential equation: $(\text{x}^{2} - 1 ) \frac{\text{dy}}{\text{dx}} + 2 \text{xy} = \frac{2}{\text{x}^{2} - 1 }.$
AnswerGiven differential equation can be written as
$\frac{\text{dy}}{\text{dx}} + \frac{2\text{x}}{\text{x}^{2} - 1 }\text{y} = \frac{2}{(\text{x}^{2} - 1 )^{2}}$
Integrating factor = $\text{e}^{\int\frac{2\text{x}}{\text{x}^{2} - 1}\text{dx}} = \text{e}^{\log(\text{x}^{2} - 1 )} = \text{x}^{2} - 1 $
$\therefore\text{ Solution is }\text{y}.(\text{x}^{2} - 1 ) =\int\frac{2}{(\text{x}^{2} - 1 )^{2}}.(\text{x}^{2} - 1 )\text{dx} + \text{c}$
$\Rightarrow\text{y}(\text{x}^{2} - 1 ) = 2 \int\frac{1}{\text{x}^{2} - 1}\text{ dx} + \text{c} $
$\Rightarrow\text{y}(\text{x}^{2} - 1 ) = \log\bigg|\frac{\text{x} - 1}{\text{x} + 1 }\bigg| + \text{c}.$
View full question & answer→Question 524 Marks
Solve the differential equation $x\frac{\text{dy}}{\text{d}x} + \text{y} = x \cos x + \sin x,$ given that y = 1 when $x = \frac{\pi}{2}.$
AnswerThe given equation can be written as
$\frac{\text{dy}}{\text{dx}} + \frac{\text{y}}{\text{x}} = \cos \text{x} + \frac{\sin \text{x}}{\text{x}}$
$\text{I.F.} = \text{e}^{\int \frac{1}{\text{x}} \text{dx}} = \text{e}^{\log \text{x}} = \text{x}$
$\therefore$ Solution is
$\text{y. x} = \int \text{(x} \cos \text{x} + \sin \text{x}) \text{dx + c}$
$\Rightarrow \text{x . y} = \text{x } \sin \text{x + c}$
$\text{or} \text{ y} = \sin \text{x} + \frac{\text{c}}{\text{x}}$
$\text{when x} = \frac{\pi}{2}, \text{y} = 1, \text{we get c = 0}$
Required solution is $\text{y} = \sin \text{x}$
View full question & answer→Question 534 Marks
Find the general solution of the following differential equation:
$( 1 + \text{y}^{2}) + x - e^{\tan-1}\text{y}) \frac{\text{dy}}{dx} = 0$
AnswerGiven differential equation can be written as
$\frac{\text{dx}}{\text{dy}} + \frac{1}{1 + \text{y}^{2}}\text{x} = \frac{e^{\tan-1}\text{y}}{1 + \text{y}^{2}}$
Integrating factor is $e^{\tan^{-1}}\text{y}$
$\therefore \text{Solution is x}.e^{\tan -1}\text{y} = \int\text{e}^{2\tan^{-1}\text{y}}\frac{1}{1 + \text{y}^{2}}\text{dy}$
$\therefore\text{x}e^{\tan -1}\text{y} = \frac{1}{2}e^{\tan-1}\text{y}\text{+C}$
View full question & answer→Question 544 Marks
Solve the differential equation:
$(\tan^{-1}\text{y} - x) \text{dy} = ( 1 + \text{y}^{2}) \text{dx}$
AnswerGiven differential equation can be writen as
$\frac{\text{dx}}{\text{dy}} + \frac{1}{1 + \text{y}^{2}} . \text{x} = \frac{\tan^{-1}\text{y}}{1 + \text{y}^{2}}$
$\therefore$ Intergrating factor is $e^{\tan^{-1}} \text{y}$
$\therefore$ Solution is: $\text{x}. e^{\tan^{-1}}\text{y} = \int \frac{\tan^{-1}{\text{y}}.e^{\tan^{-1}}\text{y}}{1 + \text{y}^{2}}\text{dy}$
$\Rightarrow \text{x .e}^{\tan^{-1}\text{y}} = \int \text{t}\text{ e}^{\text{t}}\text{ dt where } \tan^{-1}\text{y} = \text{t}$
$= \text{t e' - e' + c = e}^{\tan^{-1}\text{y}} (\tan^{-1}\text{y} - 1) + \text{c}$
$\text{or x} = \tan^{-1}\text{y} - 1 + \text{c e}^{-\tan^{-1}}\text{y}$
View full question & answer→Question 554 Marks
Find the particular solution of the differential equation$\text{e}^{x}\sqrt{1 - \text{y}^{2}} \text{ dx} + \frac{\text{y}}{\text{x}}\text{dy } = 0 $ given that y= 1 when x=0.
Answer$\text{e}^{x}\sqrt{1 - \text{y}^{2}}\text{ dx} = \frac{-\text{y}}{\text{x}}\text{ dy } \Rightarrow\text{xe}^{x}\text{dx} = \frac{-\text{y}}{\sqrt{1 - \text{y}^{2}}}\text{ dy }$
Integrating both sides
$\int\text{xe}^{x}\text{ dx} = \frac{1}{2}\int-\frac{-2\text{y}}{\sqrt{1- \text{y}^{2}}}\text{ dy}$
$\Rightarrow\text{xe}^{x} - \text{e}^{x} =\sqrt{1 - \text{y}^{2}}+\text{c}$
For x = 0, y = 1, c = – 1 $\therefore\text{ solution is: } \text{ e}^{x} (\text{x} - 1 ) = \sqrt{1 - \text{y}^{2}} - 1 .$
View full question & answer→Question 564 Marks
Show that the differential equation 2yex/y dx + (y - 2xex/y ) dy = 0 is homogeneous. Find the particular solution of this differential equation, given that x = 0 when y = 1.
AnswerGiven: 2y. ex/y dx +(y – 2x ex/y ) dy = 0
$\Rightarrow\frac{\text{dx}}{\text{dy}} = - \frac{\text{y} - 2\text{xe}^{\text{x/y}}}{2\text{y}.\text{e}^{x/y}}\Rightarrow\frac{\text{dx}}{\text{dy}} =\frac{2\text{xe}^{x/y} - \text{y}}{2\text{y.e}^{x/y}}$
Let $\text{F}(\text{x,y}) = \frac{2\text{x.e}^{x/y} - \text{y}}{2\text{y.e}^{x/y}}$
$\therefore\text{F}(\lambda\text{x},\lambda\text{y}) = \frac{2\lambda\text{x.e}^{\lambda\text{x}/\lambda\text{y}} - \lambda\text{y}}{2\lambda\text{y.e}^{\lambda\text{ x}/\lambda\text{ y}}} = \lambda^{0}\frac{2\text{xe}^{x/y} - \text{y}}{2\text{y.e}^{x/y}} = \lambda^{0}.\text{F}(\text{x,y})$
Hence, given differential equation is homogeneous.
Now, $\frac{\text{dx}}{\text{dy}} = \frac{2\text{x.e}^{x/y} - \text{y}}{2\text{y.e}^{x/y}}$ - - - - - - (i)
Let x =vy $\Rightarrow\frac{\text{dx}}{\text{dy}} = v + \text{y}.\frac{\text{d}v}{\text{dy}}$
$\therefore\text{(i)}\Rightarrow v + \text{y}.\frac{\text{d}v}{\text{dy}} = \frac{2\text{vy}.\text{e}^{\frac{\text{vy} }{\text{y}} }- \text{y}}{2\text{y.e}^{\frac{\text{vy}}{\text{y}}}}$
$\Rightarrow\text{y}.\frac{\text{dv}}{\text{dy}} =\frac{\text{y}(2v\text{e}^{v} - 1 )}{2\text{y.e}^{v}} - v\Rightarrow\text{y}.\frac{\text{d}v}{\text{dy}} = \frac{2v.\text{e}^{v} - 1}{2\text{e}^{v}} - {v}$
$\Rightarrow\text{y}.\frac{\text{d}v}{\text{dy}} = - \frac{1}{2\text{e}^{v}}\Rightarrow2\text{y.e}^{v}\text{d}v = -\text{dy}$
$\Rightarrow2\int\text{e}^{v}\text{d}v = -\int\frac{\text{dy}}{\text{y}}\Rightarrow2\text{e}^{v} = -\log\text{ y + C}$
$\Rightarrow2\text{e}^{\frac{\text{x}}{\text{y}}} = \log\text{y= C}$
When x = 0, y =1
$\therefore2\text{e}^{0} + \log 1 =\text{C}\text{ or } \text{C} = 2 $
Hence, the required solution is
$2\text{e}^{x/y} + \log\text{ y} = 2 \Rightarrow\log\text{ C} = 2 .$
View full question & answer→Question 574 Marks
If y = (tan–1x)2, show that $\text{(x}^{2}+\text{1})^{2}\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{2x}(\text{x}^{2}+{1)}\frac{\text{dy}}{\text{dx}}=2.$
Answer$\text{y}=(\tan^{-1}\text{x})^{2}\Rightarrow\frac{\text{dy}}{\text{dx}}=2\tan^{-1}\text{x}\cdot\frac{\text{1}}{\text{1+x}^{2}}.$
$\Rightarrow{(1}+\text{x}^{2})\frac{\text{dy}}{\text{dx}}=\text{2 tan}^{-1}\text{x}$
$\therefore\text{(1+x}^{2})\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{2x}\cdot\frac{\text{dy}}{\text{dx}}=\frac{\text{2}}{\text{1+x}^{2}}$
$\Rightarrow\text{(1+x}^{2})\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{2x}\text{ (1+x}^{2})\frac{\text{dy}}{\text{dx}}=2.$
View full question & answer→Question 584 Marks
Solve the following differential equation:
$\text{2x}^{2}\frac{\text{dy}}{\text{dx}}-\text{2 xy + y}^{2}=0$
Answer$\text{2x}^{2}\frac{\text{dy}}{\text{dx}}-\text{2 xy + y}^{2}=0\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{2 xy - y}^{2}}{\text{2x}^{2}}=\frac{\text{2}\frac{\text{y}}{\text{x}}-\frac{\text{y}^{2}}{\text{x}^{2}}}{\text{2}}$ Putting $\frac{\text{y}}{\text{x}}$ v so that y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v} + \text{x}\ \frac{\text{dv}}{\text{dx}}$
$\therefore\text{v + x }\frac{\text{dv}}{\text{dx}}=\text{v}-\frac{1}{2}\text{v}^{2}\therefore \text{x}\frac{\text{dv}}{\text{dx}}=-\frac{1}{2}\text{v}^{2}$
$\Rightarrow 2\int\frac{\text{dv}}{\text{v}^{2}} = -\int\frac{\text{dx}}{\text{x}}\Rightarrow\frac{2}{\text{v}}=\log\text{ x + c}$
$\therefore 2 \frac{\text{x}}{\text{y}}=\log\text{ x + c or y}=\frac{\text{2x}}{\text{log x + c}}.$
View full question & answer→Question 594 Marks
Find the particular solution of the following differential equation;
$\frac{\text{dx}}{\text{dy}}$ = 1 + x2 + y2 + x2y2, given that y = 1 when x = 0.
Answer$\frac{\text{dx}}{\text{dy}}$= 1 + x2 + y2 + x2y2 = (1 + x2)(1 + y2) $\Rightarrow\int\frac{\text{dy}}{\text{1+y}^{2}}=\int{\text{(1 + x}^{2})}\text{dx}$
$\Rightarrow\text{tan}^{-1}\text{y}=\text{x}+\frac{\text{x}^{3}}{3}+\text{c}$
x = 0, y = 1
$\Rightarrow$ c = $\frac{\pi}{4}$ $\therefore\text{ tan}^{-1}\text{y}=\text{x}+\frac{\text{x}^{3}}{3}+\frac{\pi}{4}$
OR y = $\text{tan}\Bigg(\frac{\pi}{4}+\text{x}\frac{\text{x}^{3}}{{3}}\Bigg)$ View full question & answer→Question 604 Marks
Solve the following differential equation:
ex tan y dx + (1 – ex) sec2y dy = 0.
AnswerGiven differential equation can be written as $\frac{\text{e}^{\text{x}}}{\text{1-e}^{\text{x}}}\text{ dx}+\frac{\sec^{2}\text{y}}{\tan\text{y}}\text{ dy}=0$ Integrating to get - log |1 - ex |+log|tan y| = log |c| log |tan y| = log|c (1-ex)
$\therefore$ tan y = c (1 - ex).
View full question & answer→Question 614 Marks
Solve the following differential equation:
$\cos^{2}\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\tan\text{x}.$
AnswerGiven differential equation can be written as $\frac{\text{dy}}{\text{dx}}+\text{sec}^{2}\text{x}\cdot\text{y}=\tan\text{x}\cdot\sec^{2}\text{x}$ $\text{I.F.}=\text{e}^{\int{\text{sec}^{2}\text{x dx}}}=\text{e}^{\tan\text{x}}$ $\therefore$ The Solution is $\text{y}\cdot\text{e}^{\text{tan x}}=\int\tan\text{x }\cdot\text{e}^{\text{tan x}}\text{sec}^{2}\text{ x}\text{ dx}$ $=\int\text{t.e}^\text{t}\text{dt},$ where tan x = 1. $\Rightarrow\text{y}\cdot\text{e}^{\text{tan x}}=$ (t - 1) et + c $\Rightarrow\text{y}\cdot\text{e}^{\text{tan x}}=$ (tan x - 1) etan x + c. Alternate Answer
y = (tan x - 1) + $\text{c}\cdot\text{c}^{\text{-tan x}}$. View full question & answer→Question 624 Marks
If x = a ( θ – sin θ ), y = a (1 + cos θ ), find $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}$.
Answer$\therefore\frac{\text{dx}}{\text{d}\theta}=\text{a(1 - cos}\theta)\text{ and }\frac{\text{dy}}{d\theta}=-\text{a sin}\theta$ $\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\sin\theta}{(1-\cos\theta)}$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\frac{\text{(1-cos}\theta)\text{(-cos}\theta)+\sin\theta(\sin\theta)}{(1-\cos\theta)^{2}}\cdot\frac{\text{d}\theta}{\text{dx}}$
= $\frac{(\text{1-cos}\theta)}{\text{(1 - cos}\theta)^{2}}\cdot\frac{1}{\text{a(1 - cos}\theta)}$
$= \frac{1}{\text{a(1- cos}\theta)^{2}}\text{ or }\frac{1}{\text{4a}}\text{cosec}^{4}\frac{\theta}{2}$
View full question & answer→Question 634 Marks
Find the particular solution of the differential equation satisfying the given conditions:
x2dy + (xy + y2) dx = 0 ; y = 1 when x = 1.
AnswerThe given differential equation can be written as $\frac{\text{dy}}{\text{dx}}+\frac{\text{xy+y}^{2}}{\text{x}^{2}}=0\text{ }\cdot\text{ }\text{Let y = vx}\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
$\Rightarrow\text{v + x}\frac{\text{dv}}{\text{dx}}+\text{(v + v}^{2})=0$
$\Rightarrow\text{x}\frac{\text{dv}}{\text{dx}}=-\text{ v (2 + v)}$
$\text{OR }\frac{\text{dv}}{\text{v(2+v)}}=-\frac{\text{dx}}{\text{x}}$
$\text{OR }\int\Bigg(\frac{1}{\text{v}}-\frac{1}{\text{2 + v}}\Bigg)\text{dx}=-2\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow\log\frac{\text{v}}{\text{v+2}}=\log\frac{\text{c}}{\text{x}^{2}}$
$\text{OR }\frac{\text{y}}{\text{y+2}}=\frac{\text{c}}{\text{x}^{2}}$
when x = 1, y = 1 $\Rightarrow\text{c}=\frac{1}{3}$
$\therefore$ The solution becomes
y + 2x = 3x2y.
View full question & answer→Question 644 Marks
Find the particular solution of the differential equation satisfying the given conditions:
$\frac{\text{dy}}{\text{dx}}$= y tan x, given that y = 1 when x = 0.
AnswerGiven differential equation can be written as
$\int\frac{\text{dy}}{\text{y}}=\int\tan\text{x dx}$
OR, log y = log sec x + c
when, x = 0, y = 1 $\Rightarrow$ c = 0
[Note : c = 1, if constant is taken as log c]
$\therefore$ log y = log sec x
OR y = sec x.
View full question & answer→Question 654 Marks
Find the general solution of the differential equation
$\text{x}\log\text{x}.\frac{\text{dy}}{\text{dx}}+\text{y}=\frac{2}{\text{x}}\cdot\log\text{x}$.
AnswerThe given differential equation can be written as
$\frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x log x}}\text{y}=\frac{2}{\text{x}^{2}}$
$\text{I.F.}=\text{e}^{\int\frac{1}{\text{x log x}}\text{dx}}=\text{e}^{\log(\log x)}=\log\text{x}$
The solution is y . log x = $\int\frac{2}{\text{x}^{2}}\cdot\log\text{x dx + c}$
OR, y . log x = 2
$\Bigg[\log\text{x}\cdot\Big(\frac{-1}{\text{x}}\Big)+\int\frac{\text{dx}}{{\text{x}}^{2}}\Bigg]+\text{c}=2\Bigg[\frac{-\log\text{x}}{\text{x}}-\frac{\text{1}}{\text{x}}\Bigg]+\text{c}$
$\Rightarrow\text{y}\cdot\log\text{x}=-\frac{2}{\text{x}}[1+\log\text{x}]+\text{c}$
View full question & answer→Question 664 Marks
Find $\frac{\text{dy}}{\text{dx}},\text{if y = sin}^{-1}[\text{x}\sqrt{1 - x}-\sqrt{x}\sqrt{1 - x^{2}}]$.
Answer$\text{y}=\sin^{-1}\text[{x}\sqrt{1-x}-\sqrt{x}\sqrt{1 - x^{2}}]$............(i) Let x = sin $\alpha\text{ and }\sqrt{x}=\sin\theta$ $\therefore$ (i) Becomes y = sin-1 $[\sin\alpha\cos\theta-\cos\alpha\sin\theta]$. $=\sin^{-1}[\sin(\alpha-\theta)]=\alpha-\theta$
$=\sin^{-1}\text{x}-\sin^{-1}\sqrt{x}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1}-x^2}-\frac{1}{2\sqrt{x}\sqrt{1-x}}$
View full question & answer→Question 674 Marks
Solve the following differential equation: $+ y = \cos x - \sin x.$
AnswerGetting integrating factor $ = e \int^{1 dx} = e^{x}$ $\therefore \text{Solution is y.} e^{x} = \int(\cos x- \sin x) e^{x} dx$
$\text{Unsing} \int\text{[f(x) +f'(x)]} e^{x} + \text{c we get }$
$y.e^{x} = \cos \text{x e}^{x} + c$
$\text{or y} = \cos x + \text{c e}^{-x} $
View full question & answer→Question 684 Marks
Find the particular solution, satisfying the given condition, for the following differential equation: $\frac{\text{dy}}{\text{dx}} - \frac{\text{y}}{\text{x}} + \text{cosec} \bigg(\frac{\text{y}}{\text{x}}\bigg) = \text {0; y = 0 when x} = 1.$
Answer$\text{Putting} \frac{\text{y}}{\text{x}} = \text{v}\Rightarrow \text{y = vx} \Rightarrow \frac{\text{dy}}{\text{dx}} = \text{v + x} \frac{\text{dv}}{\text{dx}}$ $\therefore \text{we get v + x} \frac{\text{dv}}{\text{dx}} -\text{v} + \text{cosec} \text{v} = \text{o}$
$\Rightarrow \sin \text{v dv} + \frac{\text{dx}}{\text{x}} = \text{o}$
$\Rightarrow - \cos \text{v} + \log|\text{x}| = \text{c}_{1} \text{or} \cos \frac{\text{y}}{\text{x}} = \log|\text{x}| + \text{c}$
$\text{When x = 1, y = o} \Rightarrow \text{c} = 1$
$\text{Hence the solution is} \cos \frac{y}{x} = 1 + \log|x|$
View full question & answer→Question 694 Marks
Solve the following differential equation: $\frac{\text{dy}}{\text{dx}} = \frac{\text{x}(\text{2y - x})}{\text{x}(\text{2y+ x)}}\text{If, y = 1 when x = 1} $
Answer$\frac{\text{dy}}{\text{dx}} = \frac{\text{x}(\text{2y - x})}{\text{x}(\text{2y+ x)}}$ $\text{y = vx} \Rightarrow \frac{\text{dy}}{\text{dx}} = \text{v + x} \frac{\text{dv}}{\text{dx}}$
$\therefore \text{v + x} \frac{\text{dv}}{\text{dx}} = \frac{\text{x[2v - 1]}}{\text{x[2v + 1]}}$
$\Rightarrow x \frac{\text{dv}}{\text{dx}} = \frac{\text{2v -1}}{\text{2v + 1}} \text{- v} = \frac{\text{2v - 1 - 2v}^{2} \text{-v}}{\text{2v + 1}}$
$= - \frac{\text{2v}^{2} \text{- v + 1}}{\text{2v + 1}}$
$\frac{\text{2v + 1}}{\text{2v}^{2} \text{- v + 1}} \text{dv} = - \frac{\text{dx}}{\text{x}}$
$\frac{1}{2} \frac{\text{4v - 1 + 3}}{\text{2v}^{2}\text{ - v+ 1}} \text{dv} = \frac{\text{-dx}}{\text{x}}$
$\frac{1}{2} \frac{\text{4v - 1 + 3}}{\text{2v}^{2}\text{ - v+ 1}} \text{dv} + \frac{3}{4} \frac{\text{dv}}{\text{v}^{2}- \frac{1}{2} \text{v} + \frac{1}{2}} = -\frac{\text{dx}}{\text{x}}$
$\frac{1}{2} \log | \text{2v}^{2} \text{- v + 1}| + \frac{3}{4} \times \frac{4}{\sqrt{7}} \tan^{-1} \frac{\text{v}-\frac{1}{4}}{{\frac{\sqrt{7}}{4}}} = -\log\text{x + c}$
$\frac{1}{2} \log\bigg|\frac{\text{2y}^{2} \text{- xy} + \text{x}^{2}}{\text{x}^{2}}\bigg| + \frac{3}{\sqrt{7}} \tan^{-1} \frac{\text{4y - x}}{\sqrt{7}\text{x}} = -\log \text{x + c}$
$\text{when x = 1, y = 1} \Rightarrow \text{c}= \frac{1}{2} \log 2 + \frac{3}{\sqrt{7}} \tan^{-1} \frac{3}{\sqrt{7}} $
View full question & answer→Question 704 Marks
Solve the following differential equation: $(\text{x}^{2} - \text{y}^{2}) \text{dx} + \text{2xy dy =0}$ given that $y = 1$ when $x = 1$
Answer$(\text{x}^{2} - \text{y}^{2}) \text{dx} + \text{2xy dy =0}$ $\frac{\text{dy}}{\text{dx}} = \frac{\text{y}^{2} - \text{x}^{2}}{\text{2xy}}$
This is a homogeneous differential equation
$\text{Let y = vx} \Rightarrow \frac{\text{dy}}{\text{dx}} = \text{v + x} \frac{\text{dv}}{\text{dx}}$
$= \text{v + x} \frac{\text{dv}}{\text{dx}} = \frac{\text x^{2}(\text v^{2}-1)}{2\text v \text x^{2}} = \frac{\text v^{2} - 1}{2\text v}$
$\text{x}\frac{\text{dv}}{\text{dx}} = \frac{\text{v}^{2} - 1 - 2\text{v}^{2}}{2\text v} = \frac{1 + \text{v}^{2}}{2\text{v}}$
$\Rightarrow \frac{2\text{v}}{1 + \text{v}^{2}} \text{dv} = - \frac{\text{dx}}{\text{x}}$
$\log | 1 + \text{v}^{2}| = -\log| \text{x}| + \log \text{c} = \log\frac{\text{c}}{\text{x}}$
$1 + \text v^{2} = \frac{\text{c}}{\text{x}} \Rightarrow 1 + \frac{\text{y}^{2}}{\text{x}^{2}}= \frac{\text{c}}{\text{x}}$
$\Rightarrow \text{x}^{2} + \text{y}^{2} = \text{cx}$
$\text{when x = 1, y = 1,} \Rightarrow \text{c} = 2$
$\therefore \text{x}^{2} + \text{y}^{2} = \text{2x}$
View full question & answer→Question 714 Marks
Solve the following differential equation: $\cos^{2} x \frac{dy}{dx} + y = \tan x$
AnswerThe given differential equation can be written as $\frac{\text{dy}}{\text{dx}} + \sec^{2} \text{x y} = \tan \text{x}.\sec^{2}\text{x}$
$\text{I.F} = \text{e}^{\int\text{pdx}}=\text{e}^{\int\sec^2\text{x dx}}=e^{\tan\text{x}}$
$\therefore$
The solution is $\text{y} .e^{\tan{\text{x}}} = \int e^{\tan\text{x}} . \tan\text{x}.\sec^{2}\text{x dx + c}$
$\text{Let} \tan{\text{x}} = \text{z} \Rightarrow \sec^{2}\text{x dx = dz}$
$\therefore \int e^{\tan \text{x}} \tan \text{x} \sec^{2}\text{x dx} = \int {\text{z e}^{\text{z}} } \text{dz + c} $
$= \text{z}.e^{\text{z}} - e^{\text{z}} + \text{c} = e^{\text{z}} (\text(z - 1) + \text{c}$
$\text{y e}^{\tan\text{x}} = e^{\tan\text{x}} ( \tan{\text{x}} - 1) + \text{c}e^{-\tan\text{x}} $
View full question & answer→Question 724 Marks
Find the particular solution of the differential equation $\text{e}^\text{x}\tan\text{y dx}+(2-\text{e}^\text{x})\text{sec}^2\text{y dy}=0,$ given that $\text{y}=\frac{\pi}{4}\ \text{x} = 0.$
Answer$\text{e}^\text{x}\tan\text{y dx}+(2-\text{e}^\text{x})\text{sec}^2\text{y dy}=0$
$\text{e}^\text{x}\tan\text{y dx}+(\text{e}^\text{x}-2)\text{sec}^2\text{y dy}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^\text{x}\tan\text{y}}{\text{e}^\text{x}\text{sec}^2\text{y}-2\text{sec}^2\text{y}}$
$\frac{\text{dx}}{\text{dy}}=\frac{\text{e}^\text{x}\text{sec}^2\text{y}-2\text{sec}^2\text{y}}{\text{e}^\text{x}\tan\text{y}}$
$\frac{\text{dx}}{\text{dy}}=\frac{\text{sec}^2\text{y}}{\tan\text{y}}-\frac{2\text{sec}^2\text{y}}{\tan\text{y}}\text{e}^{\text{-x}}$
$\frac{\text{dx}}{\text{dy}}=\frac{\text{sec}^2\text{y}}{\tan\text{y}}\big[1-2\text{e}^{-\text{x}}\big]$
$\int\frac{\text{sec}^2\text{y}}{\tan\text{y}}\text{dy}=\int\frac{1}{1-2\text{e}^{-\text{x}}}\text{dx}$
$\tan\text{y}=\text{t}$
$\text{sec}^2\text{y dy}=\text{dt}$
$\int\frac{\text{dt}}{\text{t}}=\int\frac{\text{e}^\text{x}}{\text{e}^\text{x}-2}\text{dx}$
$\text{e}^\text{x}-2=\text{u}$
$\text{e}^\text{x}\text{dx}=\text{du}$
$\log\text{t}=\log\text{u}+\log\text{C}$
$\log(\tan\text{y})=\log(\text{e}^\text{x}-2)\text{C}$
$\tan\text{y}=\text{C}(\text{e}^\text{x}-2)$
$\text{Put y }=\frac{\pi}{4},\ \text{x}=0\ \ \ \tan\frac{\pi}{4}=\text{C}(1-2)$
$\text{C}=-1$
$\tan\text{y}=-(\text{e}^\text{x}-2)$
View full question & answer→Question 734 Marks
$\text{If}\ \text{y}=\sin(\sin\text{x}),\ \text{prove that}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\tan\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\cos^2\text{x}=0.$
Answer$\text{y}=\sin(\sin\text{x})$
$\frac{\text{dy}}{\text{dx}}=\cos(\sin\text{x})\cos\text{x}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\cos(\sin\text{x})(-\sin\text{x})+\cos^2\text{x}[-\sin(\sin\text{x})]$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\sin\text{x}\cos(\sin\text{x})-\cos^2\text{x}\sin(\sin\text{x})$
$\text{L.H.S}=\frac{\text{d}^2\text{y}}{\text{dx}^2}+\tan\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\cos^2\text{x}$
$=-\sin\text{x}\cos(\sin\text{x})-\cos^2\text{x}\sin(\sin\text{x})\\+\tan\text{x}\cos\text{x}\cos(\sin\text{x})+\cos^2\text{x}\sin\text{x}(\sin\text{x})$
$=-\sin\text{x}\cos(\sin\text{x})+\frac{\sin\text{x}}{\cos\text{x}}\cos\text{x}\cos(\sin\text{x})$
$=-\sin\text{x}\cos(\sin\text{x})+\sin\text{x}\cos(\sin\text{x})$
$=0=\text{R.H.S.}$
View full question & answer→Question 744 Marks
Find the particular solution of the differential equation $\frac{\text{dy}}{\text{dx}}+2\text{y}\ \tan\text{x}=\sin\text{x},$ given that $\text{y}=0\ \text{when x}=\frac{\pi}{3}.$
Answer$\frac{\text{dy}}{\text{dx}}+2\text{y}\ \tan\text{x}=\sin\text{x}$
Differential equation is of the form
$\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q}$
$\frac{\text{dy}}{\text{dx}}+2\text{y}\ \tan\text{x}=\sin\text{x}\ \ ....(1 )$
$\text{Where P}=2\tan\text{x}\ \&\ \text{Q}=\sin\text{x}$
$\text{IF}=\text{e}^{\int\text{p dx}}$
$\text{IF}=\text{e}^{\int2\tan\text{x dx}}$
$\text{IF}=\text{e}^{2\log\sec\text{x}}$
$\text{IF}=\text{e}^{\log\sec^2\text{x}}$
$\text{IF}=\sec^2\text{x}$
$\text{y}(\text{IF})=\int(\text{Q}\times\text{IF})\text{dx}+\text{c}$
$\text{y}(\sec^2\text{x})=\int\sin\text{x}\sec^2\text{x dx}+\text{c}$
$\text{y}\sec^2\text{x}=\int\sin\text{x}\frac{1}{\cos^2\text{x}}\text{dx}+\text{C}$
$\text{y}\sec^2\text{x}=\int\frac{\sin\text{x}}{\cos\text{x}}\times\frac{1}{\cos\text{x}}\text{dx}+\text{C}$
$\text{y}\sec^2\text{x}=\int\tan\text{x}\sec\text{x dx}+\text{C}$
$\text{y}\sec^2\text{x}=\sec\text{x}+\text{C}$
$\text{y}=\frac{\sec\text{x}}{\sec^2\text{x}}+\frac{\text{c}}{\sec^2\text{x}}$
$\text{y}=\cos\text{x}+\text{C}\cos^2\text{x}\ \ ....(2)$
$\text{Putting x}=\frac{\pi}{3}\ \&\ \text{y}=0$
$0=\cos\frac{\pi}{3}+\text{C}\cos^2\frac{\pi}{3}$
$0=\frac{1}{2}+\text{C}\Big(\frac{1}{4}\Big)^2$
$\frac{-1}{2}=\text{C}\Big(\frac{1}{4}\Big)$
$\frac{-4}{2}=\text{C}$
$\text{C}=-2$
Putting value of C in (1)
$\text{y}=\cos\text{x}+\text{C}\cos^2\text{x}$
$\text{y}=\cos\text{x}-2\cos^2\text{x}$
View full question & answer→Question 754 Marks
Solve the differential equation: $\frac{\text{dy}}{\text{dx}}-\frac{2\text{x}}{1+\text{x}^2}\text{y}=\text{x}^2+2$
AnswerThe given differential equation is
$\frac{\text{dy}}{\text{dx}}-\frac{2\text{x}}{1+\text{x}^2}\text{y}=\text{x}^2+2$
This equation is of the form $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$
where $\text{P}=\frac{-2\text{x}}{1+\text{x}^2}$ and $\text{Q}=\text{x}^2+2$
Now, $\text{l.F}=\text{e}^{\int\frac{2\text{x}}{1+\text{x}^2}\text{dx}}$
$=\text{e}^{-\log\Big(\frac{1}{1+\text{x}^2}\Big)}=\frac{1}{1+\text{x}^2}$
The general solution of the given differential equation is
$\text{y}\times\text{l.F}.=\int(\text{Q}\times\text{l.F.})\text{dx}+\text{C},$ where C is an aribatry contant
$\Rightarrow\frac{\text{y}}{1+\text{x}^2}=\int\frac{\text{x}^2+2}{1+\text{x}^2}\text{dx}+\text{C}$
$=\int\Big(1+\frac{1}{\text{x}^2+1}\Big)\text{dx}+\text{C}$
$=\int\text{dx}+\int\frac{1}{\text{x}^2+1}\text{dx}+\text{C}$
$=\text{x}+\tan^{-1}\text{x}+\text{C}$
$\text{y}=(1+\text{x}^2)(\text{x}+\tan^{-1})\text{x}+\text{C}$
View full question & answer→Question 764 Marks
Solve the differential equation: $(\text{x}+1)\frac{\text{dy}}{\text{dx}}=2\text{e}^{-\text{y}}-1;\ \text{y(0)}=0.$
Answer$(\text{x}+1)\frac{\text{dy}}{\text{dx}}=2\text{e}^{-\text{y}}-1$
$\Rightarrow\frac{\text{dy}}{2\text{e}^{-\text{y}}-1}=\frac{\text{dx}}{\text{x}+1}$
$\Rightarrow\frac{\text{e}^\text{y}\text{dy}}{2-\text{e}^\text{y}}=\frac{\text{dx}}{\text{x}+1}$
Integrating both sides, we get:
$\int\frac{\text{e}^\text{y}\text{dy}}{2-\text{e}^\text{y}}=\log|\text{x}+1|+\log\text{C}\ \dots(1)$
Let $2-\text{e}^\text{y}=\text{t}.$
$\therefore\frac{\text{d}}{\text{dy}}(2-\text{e}^{\text{y}})=\frac{\text{dt}}{\text{dy}}$
$\Rightarrow-\text{e}^\text{y}=\frac{\text{dt}}{\text{dy}}$
$\Rightarrow\text{e}^\text{y}\text{dy}=-\text{dt}$
Substituting this value in equation (1), we get
$\int\frac{-\text{dt}}{\text{t}}=\log|\text{x}+1|+\log\text{C}$
$\Rightarrow-\log|\text{t}|=\log|\text{C}(\text{x}+1)|$
$\Rightarrow-\log|2-\text{e}^\text{y}|=\log|\text{C}(\text{x}+1)|$
$\Rightarrow\frac{1}{2-\text{e}^\text{y}}=\text{C}(\text{x}+1)$
$\Rightarrow2-\text{e}^\text{y}=\frac{1}{\text{C}(\text{x}+1)}\ \dots(2)$
Now, at x = 0 and y = 0, equation (2) becames:
$\Rightarrow2-1=\frac{1}{\text{C}}$
$\Rightarrow\text{c}=1$
Substituting C = 1 in equation (2) we get:
$2-\text{e}^\text{y}=\frac{1}{\text{x}+1}$
$\Rightarrow\text{e}^\text{y}=2-\frac{1}{\text{x}+1}$
$\Rightarrow\text{e}^\text{y}=\frac{2\text{x}+2-1}{\text{x}+1}$
$\Rightarrow\text{e}^\text{y}=\frac{2\text{x}+1}{\text{x}+1}$
$\Rightarrow\text{y}=\log\Big|\frac{2\text{x}+1}{\text{x}+1}\Big|,(\text{x}\neq-1)$
This is the required particular solution of the given differential equaion.
View full question & answer→Question 774 Marks
Solve the differential equation: $\text{xdy}-\text{ydx}=\sqrt{\text{x}^2+\text{y}^2}\text{ dx},$ given that $\text{y}=0$ when $\text{x}=1.$
Answer$\text{xdy}-\text{ydx}=\sqrt{\text{x}^2+\text{y}^2}\text{ dx}$
$\Rightarrow\text{xdy}=\Big[\text{y}+\sqrt{\text{x}^2+\text{y}^2}\Big]\text{ dx}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}+\sqrt{\text{x}^2+\text{y}^2}}{\text{x}}\ \dots(1)$
Let $\text{F}(\text{x, y})=\frac{\text{y}+\sqrt{\text{x}^2+\text{y}^2}}{\text{x}}$
$\therefore\text{F}(\lambda\text{x},\lambda\text{y})=\frac{\lambda\text{x}\sqrt{(\lambda\text{x})^2+(\lambda\text{y}^2)}}{\lambda\text{x}}=\frac{\text{y}+\sqrt{\text{x}^2+\text{y}^2}}{\text{x}}=\lambda^0.\text{F}(\text{x},\ \text{y})$
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
$\text{y}=\text{vx}$
$\Rightarrow\frac{\text{d}}{\text{dx}}(\text{y})=\frac{\text{d}}{\text{dx}}(\text{vx})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}$
Substitution the values of v and $\frac{\text{dy}}{\text{dx}}$ in equation (1), we get
$\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}+\sqrt{\text{x}^2+(\text{vx})^2}}{\text{x}}$
$\Rightarrow\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}=\text{v}+\sqrt{1+\text{v}^2}$
$\Rightarrow\frac{\text{dv}}{\sqrt{1+\text{v}^2}}=\frac{\text{dx}}{\text{x}}$
Integrating both sides, we get:
$\log\Big|\text{v}+\sqrt{1+\text{v}^2}\Big|=\log|\text{x}|+\log\text{C}$
$\Rightarrow\log\Bigg|\frac{\text{y}}{\text{x}}+\sqrt{1+\frac{\text{y}^2}{\text{x}^2}}\Bigg|=\log|\text{Cx}|$
$\Rightarrow\log\Bigg|\frac{\text{y}+\sqrt{\text{x}^2+\text{y}^2}}{\text{x}}\Bigg|=\log|\text{Cx}|$
$\Rightarrow\text{y}+\sqrt{\text{x}^2+\text{y}^2}=\text{Cx}^2$
This is the required solution of the given differential equation.
View full question & answer→Question 784 Marks
Solve the differential equation: $(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}+2\text{xy}-4\text{x}^2+0,$ subject to the initial condition $\text{y}(0)=0.$
AnswerThe given differential equation can be written as:
$\frac{\text{dy}}{\text{dx}}+\frac{2\text{x}}{1+\text{x}^2}\text{y}=\frac{4\text{x}^2}{1+\text{x}^2}\ \dots(1)$
This is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
$\text{P}=\frac{2\text{x}}{1+\text{x}^2}$ and $\text{Q}=\frac{4\text{x}}{1+\text{x}^2}$
$\text{I.F}=\text{e}^{{\int}\text{Pdx}}=\text{e}^{{\int}\frac{2\text{x}}{1+\text{x}^2}\text{dx}}=\text{e}^{{\log}(1+\text{x}^2)}=1+\text{x}^2$
Multipying both sides of (1) by $\text{I.F}.=(1+\text{x}^2),$ we get
$(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}+2\text{xy}=4\text{x}^2$
Integrating both sides with respect to x, we get
$\text{y}(1+\text{x}^2)=\int4\text{x}^2\text{dx}+\text{C}$
$\text{y}(1+\text{x}^2)=\frac{4\text{x}^3}{3}+\text{C}\ \dots(2)$
Given $\text{y}=0,$ when $\text{x}=0$
Substituting $\text{x}=0$ and $\text{y}=0$ in (1), we get
$0=0+\text{C}\Rightarrow\text{C}=0$
Substituting $\text{C}=0$ in (2), we get $\text{y}=\frac{4\text{x}^3}{3(1+\text{x}^2)},$ which is the required solution.
View full question & answer→Question 794 Marks
For the following differntial equations verify that the accompanying function is a solution:
| Differential equation | Function |
| $\text{y}=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$ | $\text{y}=\frac{1}{4}(\text{x}\pm\text{a})^2$ |
AnswerWe have $\text{y}=\frac{1}{4}(\text{x}\pm\text{a})^2\ ...(1)$ Differentiating both sides of (1) with respect to x, we get $\frac{\text{dy}}{\text{dx}}=\frac{1}{4}\times2(\text{x}\pm\text{a})$ $\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2}(\text{x}\pm\text{a})$ Squaring both sides we get $\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big) ^2=\Big[\frac{1}{2}(\text{x}\pm\text{a})\Big]^2$ $\Rightarrow\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)=\frac{1}{4}(\text{x}\pm\text{a})^2$ $\Rightarrow\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)^2=\text{y}$ $\therefore\text{y}=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$ Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 804 Marks
Solve the following differential equation:
$(\text{x + y})(\text{dx}-\text{dy})=\text{dx + dy}$
AnswerWe have,
$(\text{x + y})(\text{dx}-\text{dy})=\text{dx + dy}$
$\Rightarrow\text{x dx + y dx}-\text{x dy}-\text{y dy}=\text{dx + dy}$
$\Rightarrow(\text{x + y}-1)\text{dx}=(\text{x + y}+1)\text{dy}$
$\Rightarrow \frac{\text{dy}}{\text{dx}} = \frac{\text{x}+\text{y}-1}{\text{x}+\text{y}+1}$
Let $\text{ x} + \text{y} = \text{v}$
$\therefore 1+ \frac{\text{dy}}{\text{dx}} = \frac{\text{dv}}{\text{dx}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}} = \frac{\text{dv}}{\text{dx}} - 1$
$\therefore\frac{\text{dv}}{\text{dx}}-1 = \frac{\text{v}-1}{\text{v}+1}$
$\Rightarrow \frac{\text{dv}}{\text{dx}} = \frac{\text{v}-1}{\text{v}+1}+1$
$\Rightarrow \frac{\text{dv}}{\text{dx}} = \frac{\text{v}-1+\text{v}+1}{\text{v}+1}$
$\Rightarrow \frac{\text{dv}}{\text{dx}} = \frac{2\text{v}}{\text{v}+1}$
$\Rightarrow \frac{\text{v}+1}{2\text{v}}\text{dv} = \text{dx}$
Integrating both sides, we get
$\int \frac{\text{v}+1}{2\text{v}}\text{dv} = \int\text{dx}$
$\Rightarrow \frac{1}{2}\int\text{dv}+\frac{1}{2}\int\frac{1}{\text{v}}\text{dv} = \int\text{dx}$
$\Rightarrow \frac{1}{2}\text{v}+\frac{1}{2}\log|\text{v}| = \text{x}+\text{C}$
$\Rightarrow \frac{1}{2}(\text{x}+\text{y})+\frac{1}{2}\log|\text{x}+\text{y}| = \text{x}+\text{C}$
$\Rightarrow \frac{1}{2}(\text{y}-\text{x})+\frac{1}{2}\log|\text{x}+\text{y}| = \text{C}$
View full question & answer→Question 814 Marks
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\sin\Big(\frac{\text{y}}{\text{x}}\Big)$
Answer$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\sin\Big(\frac{\text{y}}{\text{x}}\Big)$
This is a homogeneous differential equation.
Putting y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\text{v}+\sin\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\text{v}+\sin\text{v}-\text{v}$
$\Rightarrow\ \frac{1}{\sin\text{v}}\text{dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{\sin\text{v}}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\text{cosec v dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \log\Big|\tan\frac{\text{v}}{2}\Big|=\log|\text{x}|+\log\text{C}$
$\Rightarrow\ \log\Big|\tan\frac{\text{v}}{2}\Big|=\log|\text{Cx}|$
$\Rightarrow\ \tan\frac{\text{v}}{2}=\text{Cx}$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\Rightarrow\ \tan\Big(\frac{\text{y}}{2\text{x}}\Big)=\text{Cx}$
Hence, $\tan\Big(\frac{\text{y}}{2\text{x}}\Big)=\text{Cx}$ is the required solution.
View full question & answer→Question 824 Marks
Find the particular solution of the differential equation $(\text{x}-\text{y})\frac{\text{dy}}{\text{dx}}=\text{x +2y},$ given that when x = 1, y = 0.
AnswerConsider the given equation $(\text{x}-\text{y})\frac{\text{dy}}{\text{dx}}=\text{x +2y}$ This is a homogeneous equation. Substituting y = vx and $\frac{\text{dy}}{\text{dx}}=\Big(\text{v + x}\frac{\text{dv}}{\text{dx}}\Big)$ In the above equation, we have,
$(\text{x}-\text{vx})\Big(\text{v + x}\frac{\text{dv}}{\text{dx}}\Big)=\text{x +2vx}$ $\Rightarrow\ (1-\text{v})\Big(\text{v + x}\frac{\text{dv}}{\text{dx}}\Big)=1+2\text{v}$ $\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{1+2\text{v}}{1-\text{v}}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+2\text{v}}{1-\text{v}}-\text{v}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+2\text{v}-\text{v}(1-\text{v})}{1-\text{v}}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+2\text{v}-\text{v}+\text{v}^2}{1-\text{v}}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}+\text{v}^2}{1-\text{v}}$ $\Rightarrow\ \frac{(1-\text{v})\text{dv}}{(1+\text{v}+\text{v}^2)}=\frac{\text{dx}}{\text{x}}$ Integrating both sides, we have, $\Rightarrow\ \int\frac{(1-\text{v})\text{dv}}{(1+\text{v}+\text{v}^2)}=\int\frac{\text{dx}}{\text{x}}$ $\Rightarrow\ \frac{3}2\int\frac{\text{dv}}{(1+\text{v}+\text{v}^2)}-\int\frac{1}2\frac{(2\text{v}+1)\text{dv}}{(1+\text{v}+\text{v}^2)}=\int\frac{\text{dx}}{\text{x}}$ $\Rightarrow\ \frac{3}2\int\frac{\text{dv}}{\text{v}^2+\frac{1}4+\text{v}+\frac{3}4}-\int\frac{1}2\frac{(2\text{v}+1)\text{dv}}{(1+\text{v}+\text{v}^2)}=\int\frac{\text{dx}}{\text{x}}$ $\Rightarrow\ \frac{3}2\int\frac{\text{dv}}{\big(\text{v}+\frac{1}2\big)^2+\big(\frac{\sqrt3}2\big)^2}-\int\frac{1}2\frac{(2\text{v}+1)\text{dv}}{(1+\text{v}+\text{v}^2)}=\int\frac{\text{dx}}{\text{x}}$ $\Rightarrow\ \frac{3}2\times\frac{1}{\frac{\sqrt3}{2}}\tan^{-1}\frac{\text{v}+\frac{1}{2}}{\frac{\sqrt3}{2}}-\frac{1}2\log(1+\text{v}+\text{v}^2)=\log\text{x}+\text{C}$ $\Rightarrow\ \sqrt3\tan^{-1}\frac{2\text{v}+1}{\sqrt3}-\frac{1}2\log(1+\text{v}+\text{v}^2)=\log\text{x}+\text{C}$ $\Rightarrow\ \sqrt3\tan^{-1}\frac{2\big(\frac{\text{y}}{\text{x}}\big)+1}{\sqrt3}-\frac{1}{2}\log\Big(1+\Big(\frac{\text{y}}{\text{x}}\Big)+\Big(\frac{\text{y}}{\text{x}}\Big)^2\Big)=\log\text{x}+\text{C}\ \dots(1)$ Given that when x = 1, y = 0 Substituting the values, in the above equation, we get, $\Rightarrow\ \sqrt3\tan^{-1}\frac{2\times0+1}{\sqrt3}-\frac{1}2\log(1+0+0^2)=\log1+\text{C}$ $\Rightarrow\ \sqrt3\tan^{-1}\frac{1}{\sqrt3}-\frac{1}{2}\times0=0+\text{C}$ $\Rightarrow\ \text{C}=\sqrt3\times\frac{\pi}{6}$ $\Rightarrow\ \text{C}=\frac{\pi}{2\sqrt3}$ Thus equation (1) becomes, $\sqrt3\tan^{-1}\frac{2\big(\frac{\text{y}}{\text{x}}\big)+1}{\sqrt3}-\frac{1}2\log\Big(1+\Big(\frac{\text{y}}{\text{x}}\Big)+\Big(\frac{\text{y}}{\text{x}}\Big)^2\Big)=\log\text{x}+\frac{\pi}{2\sqrt3}$ $\Rightarrow\ \sqrt3\tan^{-1}\frac{2\text{y + x}}{\text{x}\sqrt3}-\frac{\pi}{2\sqrt3}=\log\text{x}+\frac{1}2\log\Big(1+\Big(\frac{\text{y}}{\text{x}}\Big)+\Big(\frac{\text{y}}{\text{x}}\Big)^2\Big)$ $\Rightarrow\ 2\sqrt3\tan^{-1}\frac{2\text{y + x}}{\text{x}\sqrt3}-\frac{\pi}{\sqrt3}=\log\text{x}^2+\log\Big(\frac{\text{x}^2+\text{xy}+\text{y}^2}{\text{x}^2}\Big)$ $\Rightarrow\ 2\sqrt3\tan^{-1}\frac{2\text{y + x}}{\text{x}\sqrt3}-\frac{\pi}{\sqrt3}=\log(\text{x}^2+\text{xy}+\text{y}^2)$ View full question & answer→Question 834 Marks
Solve the following differential equations:
$\text{x}\frac{\text{dy}}{\text{dx}}+\cot\text{y}=0,$ given that $\text{y}=\frac{\pi}{4},$ when $\text{x}=\sqrt{2}.$
AnswerWe have,
$\text{x}\frac{\text{dy}}{\text{dx}}+\cot\text{y}=0$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}=-\cot\text{y}$
$\Rightarrow\tan\text{y dy}=-\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\tan\text{y dy}=-\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\log|\sec\text{y}|=-\log|\text{x}|+\log\text{C}$
$\Rightarrow\log(|\text{x}||\sec\text{y}|)=\log\text{C}$
$\Rightarrow\text{x}\sec\text{y = C}...(1)$
Given: $\text{x}=\sqrt{2},\text{y}=\frac{\pi}{4}.$
Substituting the values of x and y in (1), we get
$\sqrt{2}\sec\frac{\pi}{4}=\text{C}$
$\Rightarrow\text{C}=2$
Substituting the value of C in (1), we get
$\text{x}\sec\text{y}=2$
$\Rightarrow\text{x}=2\cos\text{y}$
Hence, $\text{x}=2\cos\text{y}$ is the required solution.
View full question & answer→Question 844 Marks
The normal to a given curve at each point (x, y) on the curve passes through the point (3, 0). If the curve contains the point (3, 4), find its equation.
AnswerEquation of normal on point (x, y) on the curve
$\text{y}-\text{y}=\frac{-\text{dx}}{\text{dy}}(\text{x}-\text{x})$
Its passing through (3, 0)
$\Rightarrow\text{0}-\text{y}=\frac{-\text{dx}}{\text{dy}}(3-\text{x})$
$\Rightarrow \text{y}=\frac{\text{dx}}{\text{dy}}(3-\text{x})$
$\Rightarrow \text{y}\ \text{dy}=(3-\text{x})\text{dx}$
$\Rightarrow \int\text{y}\ \text{dy}=\int(3-\text{x})\text{dx}$
$\Rightarrow \frac{\text{y}^{2}}{2}=3\text{x}-\frac{\text{x}^{2}}{2}+\text{C}\ ...(\text{i})$
It passing through (3, 4),
$\frac{16}{2}=9-\frac{9}{2}+\text{C}$
$\frac{16}{2}=\frac{9}{2}+\text{C}$
$\text{C}=7$
Put $\text{C}=7$ is equation (i)
$\frac{\text{y}^{2}}{2}=3\text{x}-\frac{\text{x}^{2}}{2}+\frac{7}{2}$
$\text{y}^{2}=6\text{x}-\text{x}^{2}+{7}$
View full question & answer→Question 854 Marks
Solve the following initial value problems:
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=\log\text{x},\text{ y}(1)=0$
AnswerWe have,
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=\log\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}=\frac{\log\text{x}}{\text{x}}\ ...(\text{1})$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=-\frac{1}{\text{x}}$ and $\text{Q}=\frac{\log\text{x}}{\text{x}}$
$\therefore\text{I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}{-\int\frac{1}{\text{x}}}\text{ dx}$
$=\text{e}^{-\log\text{x}}$
$=\frac{1}{\text{x}}$
Multiplying both sides of (1) by $\text{I.F.}=\frac{1}{\text{x}},$ we get
$\frac{1}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}}\text{y}\Big)=\frac{1}{\text{x}}\times\frac{\log\text{x}}{\text{x}}$
$\Rightarrow\frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}^2}\text{y}=\frac{\log\text{x}}{\text{x}^2}$
Integrsting both sides with respect to x, we get
$\text{y}\frac{1}{\text{x}}=\int\frac{1}{\text{x}^2}\times\log\text{x dx}+\text{C}$
$\Rightarrow\frac{\text{y}}{\text{x}}=\log\text{x}\int\frac{1}{\text{x}^2}\text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\log\text{x})\int\frac{1}{\text{x}^2}\text{dx}\Big]\text{dx}+\text{C}$
$\Rightarrow\frac{\text{y}}{\text{x}}=-\frac{\log\text{x}}{\text{x}}+\int\frac{1}{\text{x}^2}\text{dx}+\text{C}$
$\Rightarrow\frac{\text{y}}{\text{x}}=-\frac{\log\text{x}}{\text{x}}-\frac{1}{\text{x}}+\text{C}$
$\Rightarrow\text{y}=-\log\text{x}-1+\text{Cx}\ ...(\text{ii})$
Now,
$\text{y}(1)=0$
$\therefore\ 0=-0-1+\text{C}(1)$
$\Rightarrow\text{C}=1$
Putting the value of C in (2) we get
$\text{y}=-\log\text{x}-1+\text{x}$
$\Rightarrow\text{y}=\text{x}-1-\log\text{x}$
Hence, $\text{y}=\text{x}-1-\log\text{x}$ is the required solution.
View full question & answer→Question 864 Marks
The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.
AnswerLet v be volume of spherical balloon of radius r. $\therefore\ \text{v}=\frac{4}{3}\pi\text{r}^3\ ...(1)$ From give condition,
$\frac{\text{dv}}{\text{dt}}=\text{k}\ \text{or}\ \frac{\text{d}}{\text{dt}}\bigg[\frac{4}{3}\pi\text{r}^3\bigg]=\text{k}\ \ [\because \text{of}\ (1)]$ $\therefore\frac{4\pi}{3}.\ 3\ \text{r}^2\ \frac{\text{dr}}{\text{dt}}=\text{k}\ \ \text{or}\ \ 4\pi\text{r}^2\ \frac{\text{dr}}{\text{dt}}=\text{k}$
Separating the variables and integrating, we get.
$4\pi\int\text{r}^2\text{dr}=\text{k}\int\text{dt}\ \ \text{or}\ \ 4\pi \frac{\text{r}^3}{3}$ $=\text{k}\ \text{t}+\text{c}\ ...(2)$
Now t = 0 when r = 3 $\therefore\ 4\pi\frac{(3)^3}{3}=\text{k}\times0 +\text{c}\ \ \Rightarrow\ \ \text{c}=36\pi\ ...(3)$ Again t = 3 when r = 6
$\therefore\ 4\pi\frac{(3)^3}{3}(6)^3=3\ \text{k}+36\pi\ \ [\because\ \text{of}\ (3)]$
$\therefore\ 288\pi=3\text{k}+36\pi\ \text{or}\ 3\text{k}=252\pi$
$\therefore\ \text{k}=84\pi$
$\text{Putting k}=84\pi,\ \text{c}=36\pi\ \text{in (2), we get}$
$\frac{4\pi}{3}\text{r}^3=84\pi\ \text{t}+36\pi\ \text{or}\ \frac{\text{r}^3}{3}=21\ \text{t}+9$
$\therefore\ \text{r}^3=63\ \text{t}+27\ \ \Rightarrow\ \ \text{r}=[9(7 \text{t}+3)]^\frac{1}{3}$
View full question & answer→Question 874 Marks
For each of the differential equation in find the particular solution satisfying the given condition: $\text{x}^2\ \text{dy}+(\text{xy}+\text{y}^2)\ \text{dx}=0;\ \text{y}=1\ \text{when}\ \text{x}=1$
AnswerGiven: Differential equation $\text{x}^2\ \text{dy}+(\text{xy}+\text{y}^2)\ \text{dx}=0$ $\Rightarrow\ \ \text{x}^2\ \text{dy}-(\text{xy}+\text{y}^2)\ \text{dx}\ \ $ $\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=-\frac{\text{y}(\text{x}+\text{y})}{\text{x}^2}=-\frac{\text{xy}\Big(1+\frac{\text{y}}{\text{x}}\Big)}{\text{x}^2}$ $\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}\Big(1+\frac{\text{y}}{\text{x}}\Big)=f\Big(\frac{\text{y}}{\text{x}}\Big)\ \ ...\text{(i)}$ Therefore the given differential equation is homogeneous. $\text{Putting}\frac{\text{y}}{\text{x}}=\text{v}\ \ \Rightarrow\ \ \text{y}=\text{vx}\ \ $ $\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}$ $\text{Putting these values of}\ \frac{\text{y}}{\text{x}}\ \text{and}\ \frac{\text{dv}}{\text{dx}}\ \text{in eq. (ii), we have}$ $\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}=-\text{v}(1+\text{v})=-\text{v}-\text{v}^2\ \ $ $\Rightarrow\ \ \text{x}\frac{\text{dv}}{\text{dx}}=-\text{v}^2-2\text{v}$ $\Rightarrow\ \ \text{x}\frac{\text{dv}}{\text{dx}}=-\text{v}(\text{v}+2)\ \ \Rightarrow\ \ \frac{\text{dv}}{\text{v}(\text{v}+2)}=-\frac{\text{dx}}{\text{x}}$ $\text{Integrating both sides,}$ $\ \ \int\frac{1} {\text{v}(\text{v}+2)}\text{dv}=-\int\frac{1}{\text{x}}\text{dx}$ $\Rightarrow\ \ \frac{1}{2}\int\frac{2} {\text{v}(\text{v}+2)}\text{dv}=-\log|\text{x}|+\log|\text{c}|\ \ $ $\Rightarrow\ \ \frac{1}{2}\int\frac{(\text{v}+2)-\text{v}}{\text{v}(\text{v}+2)}\ \text{dv}=-\log|\text{x}|+\log|\text{c}|$ $\Rightarrow\ \ \int\Big(\frac{1}{\text{v}}-\frac{1}{\text{v+2}}\Big)\text{dv}=-2\log|\text{x}|+\log|\text{c}|$ $\Rightarrow\ \ \log\Big|\frac{\text{v}}{\text{v}+2}\Big|=\log\big|\text{x}^{-2}\big|+\log|\text{c}|$ $\Rightarrow\ \ \log\Big|\frac{\text{v}}{\text{v}+2}\Big|=\log\big|\text{cx}^{-2}\big|\ \ $ $\Rightarrow\ \ \frac{\text{v}}{\text{v}+2}=\pm\text{cx}^{-2}$ $\text{Putting}\ \text{v}=\frac{\text{y}}{\text{x}},\ \ \frac{\frac{\text{y}}{\text{x}}}{\frac{\text{y}}{\text{x}}+2}=\pm\text{cx}^{-2}\ \ $ $\Rightarrow\ \ \frac{\text{y}}{\text{y}+2\text{x}}=\pm\text{cx}^{-2}$ $\Rightarrow\ \ \text{x}^{2}\text{y}=\text{C}(\text{y}+2\text{x})\ \ \text{where C}=\pm\text{c}\ \ .....(\text{ii})$ Now putting x = 1 and y = 1 in eq. (ii), we get $1=3\text{C}\ \ \Rightarrow\ \ \text{C}=\frac{1}{3}$ Putting value of C in eq. (ii),
$\text{x}^{2}\text{y}=\frac{1}{3}(\text{y}+2\text{x})\ \ \Rightarrow\ \ 3\text{x}^2\text{y}=\text{y}+2\text{x}$
View full question & answer→Question 884 Marks
Find the particular solution of the differential equation $\frac{\text{dy}}{\text{dx}}=-4\text{xy}^2$ given that $\text{y}=1.$ when $\text{x}=0.$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=-4\text{xy}^2$
$\Rightarrow\frac{1}{\text{y}^2}\text{dy}=-4\text{x dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}^2}\text{dy}=-4\int\text{x dx}$
$\Rightarrow-\frac{1}{\text{y}}=-4\times\frac{\text{x}^2}{2}+\text{C}$
$\Rightarrow-\frac{1}{\text{y}}=-2\text{x}^2+\text{C}...(1)$
It is given that at $\text{x}=0,\text{y}=1.$
Substituting the valuse of x and y in (1), we get
$\text{C}=-1$
Therefore, substituting the value of C in (1), we get
$-\frac{1}{\text{y}}=-2\text{x}^2-1$
$\Rightarrow\text{y}=\frac{1}{2\text{x}^2+1}$
Hence, $\text{y}=\frac{1}{2\text{x}^2+1}$ is the required solution.
View full question & answer→Question 894 Marks
Solve the differential equation $(\text{y}+3\text{x}^2)\frac{\text{dx}}{\text{dy}}=\text{x}$
AnswerWe have,
$(\text{y}+3\text{x}^2)\frac{\text{dx}}{\text{dy}}=\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dy}}=\frac{\text{y}+3{\text{x}^{\text{2}}}}{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}}{\text{y}}=3{\text{x}}\ ...(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=-\frac{1}{\text{x}}$ and $\text{Q}=3\text{x}$
$\therefore \ \text{I}.\text{F}. = \text{e}^{\int{\text{P}\text{dx}}}$
$ =\text{e}^{-\int\frac{1}{\text{x}}\text{dx}}$
$=\text{e}^{-\log\text{x}}$
$=\frac{1}{\text{x}}$
Multiplying both sides of (1) by $\text{I.F.}=\frac{1}{\text{x}},$ we get
$\frac{1}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}}{\text{y}}\Big)=\frac{1}{\text{x}}3\text{x}$
$\Rightarrow\frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}^{2}}\text{y}=3$
Integrating both sides with respect to x, we get
$\frac{1}{\text{x}}\text{y}=3\int\text{dx}+\text{C}$
$\Rightarrow\frac{\text{y}}{\text{x}}=3\text{x}+\text{C}$
Hence, $\frac{\text{y}}{\text{x}}=3\text{x}+\text{C}$ is the required solution.
View full question & answer→Question 904 Marks
Solve the following differential equations: $\text{x}\sqrt{1-\text{y}^2}\text{dx}+\text{y}\sqrt{1-\text{x}^2}\text{dy}=0$
AnswerWe have,
$\text{x}\sqrt{1-\text{y}^2}\text{dx}+\text{y}\sqrt{1-\text{x}^2}\text{dy}=0$
$\Rightarrow\text{y}\sqrt{1-\text{x}^2}\text{dy}=-\text{x}\sqrt{1-\text{y}^2}\text{dx}$
$\Rightarrow\frac{\text{y}}{\sqrt{1-\text{y}^2}}\text{dy}=-\frac{\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}$
Integrating both sides, we get
$\int\frac{\text{y}}{\sqrt{1-\text{y}^2}}\text{dy}=-\int\frac{\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}$
Substituting $1-\text{y}^2=\text{t}$ and $1-\text{x}^2=\text{u},$ we get
$-2\text{y dy = dt}$ and $-2\text{x dy = du}$
$\therefore\frac{-1}{2}\int\frac{1}{\sqrt{\text{t}}}\text{dt}=\frac{1}{2}\int\frac{1}{\sqrt{\text{u}}}\text{du}$
$\Rightarrow-\text{t}^{\frac{1}{2}}=\text{u}^{\frac{1}{2}}+\text{K}$
$\Rightarrow\sqrt{1-\text{x}^2}+\sqrt{1-\text{y}^2}=-\text{K}$
$\Rightarrow\sqrt{1-\text{x}^2}+\sqrt{1-\text{y}^2}=\text{C}$ (where, C = K)
Hence, $\sqrt{1-\text{x}^2}+\sqrt{1-\text{y}^2}=\text{C}$ is the required solution.
View full question & answer→Question 914 Marks
Find one-parameter families of solution curves of the following differential equation: (or solve the following differential equation) $\frac{\text{dy}}{\text{dx}}+3\text{y}=\text{e}^{\text{mx}},$ m is given real number.
AnswerWe have, $\frac{\text{dy}}{\text{dx}}+3\text{y}=\text{e}^{\text{mx}}\ \dots(1)$ Clearly, it is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}+\text{Py = Q}$ where $\text{P}=3$ $\text{Q}=\text{e}^{\text{mx}}$ $\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$ $=\text{e}^{\int3\text{dx}}$ $=\text{e}^{3\text{x}}$ Multiplying both sides of (1) by e3x,
we get
$\text{e}^{3\text{x}}\Big(\frac{\text{dx}}{\text{dy}}+3\text{y}\Big)=\text{e}^{3\text{x}}\text{e}^{\text{mx}}$ $\Rightarrow\ \text{e}^{\text{3x}}\frac{\text{dx}}{\text{dy}}+3\text{e}^{3\text{x}}\text{y}=\text{e}^{(\text{m}+3)\text{x}}$ Integrating both sides with respect to x, we get
$\text{ye}^{\text{3x}}=\int\text{e}^{(\text{m}+3)\text{x}}\text{dx + C}$ (when
$\text{m}+3\neq0$)
$\Rightarrow\ \text{ye}^{\text{3x}}=\frac{\text{e}^{(\text{m}+3)\text{x}}}{\text{m}+3}+\text{C}$ $\Rightarrow\ \text{y}=\frac{\text{e}^{\text{mx}}}{\text{m}+3}+\text{Ce}^{-3\text{x}}$ $\text{ye}^{3\text{x}}=\int\text{e}^{0\times\text{x}}\text{dx + C}$ (when
$\text{m}+3=0$)
$\Rightarrow\ \text{ye}^{3\text{x}}=\int\text{dx + C}$ $\Rightarrow\ \text{ye}^{3\text{x}}=\text{x + C}$ $\Rightarrow\ \text{y}=(\text{x + C})\text{e}^{-3\text{x}}$ Hence,
$\text{y}=\frac{\text{e}^{\text{mx}}}{\text{m}+3}+\text{Ce}^{-3\text{x}},$ where
$\text{m}+3\neq0$ and
$\text{y}=(\text{x + C})\text{e}^{-3\text{x}},$ where
$\text{m}+3=0$ are required solutions.
View full question & answer→Question 924 Marks
Find one-parameter families of solution curves of the following differential equation: (or solve the following differential equation) $\text{x}\frac{\text{dy}}{\text{dx}}+2\text{y}=\text{x}^2\log\text{x}$
AnswerWe have $\text{x}\frac{\text{dy}}{\text{dx}}+2\text{y}=\text{x}^2\log\text{x}$
Dividing both sides by x, we get $\frac{\text{dy}}{\text{dx}}+\frac{2\text{y}}{\text{x}}=2\log\text{x}$
Comparing with
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$ we get $\text{P}=\frac{2}{\text{x}}$
$\text{Q}=\text{x}\log\text{x}$
Now, $\text{I.F}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\frac{2\text{}}{\text{x}}}\text{ dx}$
$=\text{e}^{2\log|\text{x}|}$
$=\text{x}^2$
So, the solution is given by $\text{y}\times\text{I.F.}=\int\text{Q}\times\text{I.F.}\text{ dx}+\text{C}$
$\Rightarrow\text{x}^2\text{y}=\int\text{x}^3\log\text{x dx}+\text{C}$
$\Rightarrow\text{x}^2\text{y}=\log\text{x}\int\text{x}^3\text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\log\text{x})\int\text{x}^3\text{dx}\Big]\text{dx}+\text{C}$
$\Rightarrow\text{x}^2\text{y}=\frac{\text{x}^4\log\text{x}}{4}-\int\frac{\text{x}^3}{4}\text{dx}+\text{C}$
$\Rightarrow\text{x}^2\text{y}=\frac{\text{x}^4\log\text{x}}{4}-\frac{\text{x}^3}{16}\text{dx}+\text{C}$
$\Rightarrow\text{y}=\frac{\text{x}^2\log\text{x}}{4}-\frac{\text{x}^2}{16}+\frac{\text{C}}{\text{x}^2}$
$\Rightarrow\text{y}=\frac{\text{x}^2}{16}(4\log\text{x}-1)+\frac{\text{C}}{\text{x}^2}$
View full question & answer→Question 934 Marks
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}+\frac{1+\text{y}^2}{\text{y}}=0$
AnswerWe have
$\frac{\text{dy}}{\text{dx}}+\frac{1+\text{y}^2}{\text{y}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\Big(\frac{(1+\text{y}^2}{\text{y}}\Big)$
$\Rightarrow\frac{\text{dx}}{\text{dy}}=-\frac{\text{y}}{1+\text{y}^2}$
$\Rightarrow\text{dx}=\Big(-\frac{\text{y}}{1+\text{y}^2}\Big)\text{dy}$
Integrating both sides, we get
$\int\text{dx}=\int\Big(-\frac{\text{y}}{1+\text{y}^2}\Big)\text{dy}$
$\Rightarrow\text{x}=\int\Big(-\frac{\text{y}}{1+\text{y}^2}\Big)\text{dy}$
Putting 1 + y2 = t we get
2y dy dt
$\therefore\text{x}=-\frac{1}{2}\int\frac{1}{\text{t}}\text{dt}$
$\Rightarrow\text{x}=-\frac{1}{2}\log|\text{t}|+\text{C}$
$\Rightarrow\text{x}=-\frac{1}{2}\log|1+\text{y}^2|+\text{C}$
$\Rightarrow\text{x}+\frac{1}{2}\log|1+\text{y}^2|=\text{C}$
Hence, $\text{x}+\frac{1}{2}\log|1+\text{y}^2|=\text{C}$ is the required solution.
View full question & answer→Question 944 Marks
Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=4\text{x }\text{cosec x},\text{ y}\Big(\frac{\pi}{2}\Big)=0$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=4\text{x }\text{cosec x}\ ...(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=\cot\text{x}$ and $\text{Q}=4\text{x cosec x}$
$\therefore\text{ I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\cos\text{x dx}}$
$=\text{e}^{\log|\sin\text{x}|}$
$=\sin\text{x}$
Multiplying both sides of (1) by $\text{I.F.}=\sin\text{x},$ we get
$\sin\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}\Big)=\sin\text{x}(4\text{x cosec x})$
$\Rightarrow\sin\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}\Big)=4\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\sin\text{x}=4\int\text{x dx}+\text{C}$
$\Rightarrow\text{y}\sin\text{x}=2\text{x}^2+\text{C}\ ...(2)$
Now,
$\text{y}\Big(\frac{\pi}{2}\Big)=0$
$\therefore\ 0\times\sin\Big(\frac{\pi}{2}\Big)=2\Big(\frac{\pi}{2}\Big)^2+\text{C}$
$\Rightarrow\text{C}=-\frac{\pi^2}{2}$
Putting the value of C in (2) we get
$\text{y}\sin\text{x}=2\text{x}^2-\frac{\pi^2}{2}$
Hence, $\text{y}\sin\text{x}=2\text{x}^2-\frac{\pi^2}{2}$ is the required solution.
View full question & answer→Question 954 Marks
Solve the following differential equation:
$2\text{xy dx}+(\text{x}^2+2\text{y}^2)\text{dy}=0$
AnswerHere, $2\text{xy dx}+(\text{x}^2+2\text{y}^2)\text{dy}=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{2\text{xy}}{\text{x}^2-2\text{y}^2}$
It is a homogeneous equation.
Put y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{2\text{xvx}}{\text{x}^2+2\text{v}^2\text{x}^2}$
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{2\text{v}}{1+2\text{v}^2}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{2\text{v}}{1-2\text{v}^2}-\text{v}$
$=\frac{2\text{v}-\text{v}+2\text{v}^3}{1+2\text{v}^2}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}-2\text{v}^3}{1+2\text{v}^2}$
$\int\frac{1+2\text{v}^2}{\text{v}-2\text{v}^3}\text{dv}=\int\frac{\text{dx}}{\text{x}}\ \dots(\text{i})$
$\frac{1+2\text{v}^2}{\text{v}-2\text{v}^3}=\frac{1+2\text{v}^2}{\text{v}(1-2\text{v}^2)}$
$\frac{1+2\text{v}^2}{\text{v}(1-2\text{v}^2)}=\frac{\text{A}}{\text{v}}+\frac{\text{Bv + C}}{1-2\text{v}^2}$
$\frac{1+2\text{v}^2}{\text{v}(1-2\text{v}^2)}=\frac{\text{A}(1-2\text{v}^2)+(\text{Bv + C)}\text{v}}{\text{v}(1-2\text{v}^2)}$
$1+2\text{v}^2=\text{A}-2\text{Av}^2+\text{Bv}^2+\text{Cv}$
$1+2\text{v}^2=\text{v}^2(-2\text{A + B})+\text{Cv + A}$
Comparing the co-efficients of like powers of v,
A = 1
C = 0
-2A + B = 2
-2 + B = 0
B = 4
$\frac{1+2\text{v}^2}{\text{v}-2\text{v}^3}=\frac{1}{\text{v}}+\frac{4\text{v}}{1-2\text{v}^2}$
$\frac{1+2\text{v}^2}{\text{v}-2\text{v}^3}=\frac{1}{\text{v}}-\frac{(-4\text{v})}{(1-2\text{v}^2)}$
View full question & answer→Question 964 Marks
Find the particular solution of the differential equation $\text{x}\cos\Big(\frac{\text{y}}{\text{x}}\Big)\frac{\text{dy}}{\text{dx}}=\text{y}\cos\Big(\frac{\text{y}}{\text{x}}\Big)+\text{x},$ given that when $\text{x}=1,\text{y}=\frac{\pi}4$.
Answer$\text{x}\cos\Big(\frac{\text{y}}{\text{x}}\Big)\frac{\text{dy}}{\text{dx}}=\text{y}\cos\Big(\frac{\text{y}}{\text{x}}\Big)+\text{x}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}\cos\Big(\frac{\text{y}}{\text{x}}\Big)+\text{x}}{\text{x}\cos\Big(\frac{\text{y}}{\text{x}}\Big)}$
This is a homogeneous differential equation.
puttuing y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}},$ we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}+\cos\text{v + x}}{\text{x}\cos\text{v}}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}+\cos\text{v + 1}}{\cos\text{v}}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}\cos\text{v}+1-\text{v}\cos\text{v}}{\cos{\text{v}}}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1}{\cos\text{v}}$
$\Rightarrow\ \cos\text{v dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\cos\text{v dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \sin\text{v}=\log|\text{x}|+\text{C}$
Putting $\text{v}=\frac{\text{y}}{\text{x}},$ we get
$\sin\frac{\text{y}}{\text{x}}=\log|\text{x}|+\text{C}\ \dots(1)$
At $\text{x}=1,\text{y}=\frac{\pi}4$ (Given)
Putting $\text{x}=1$ and $\text{y}=\frac{\pi}4$ in (1), we get
$\text{C}=\frac{1}{\sqrt2}$
Putting $\text{C}=\frac{1}{\sqrt2}$ in (1), we get
$\sin\frac{\text{y}}{\text{x}}=\log|\text{x}|+\frac{1}{\sqrt2}$
Hence, $\sin\frac{\text{y}}{\text{x}}=\log|\text{x}|+\frac{1}{\sqrt2}$ is the required solution.
View full question & answer→Question 974 Marks
Solve the following differential equation:
$(1+\text{y}^2)+(\text{x}-\text{e}^{\tan^{-1}\text{y}})\frac{\text{dy}}{\text{dx}}=0$
AnswerWe have, $(1+\text{y}^2)+(\text{x}-\text{e}^{\tan^{-1}\text{y}})\frac{\text{dy}}{\text{dx}}=0$ $\Rightarrow\ (\text{x}-\text{e}^{\tan^{-1}\text{y}})\frac{\text{dy}}{\text{dx}}=-(1+\text{y}^2)$ $\Rightarrow\ \frac{\text{dy}}{\text{dx}}=-\frac{(1+\text{y}^2)}{(\text{x}-\text{e}^{\tan^{-1}\text{y}})}$ $\Rightarrow\ \frac{\text{dx}}{\text{dy}}=-\frac{\text{x}-\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}$
$\Rightarrow\ \frac{\text{dx}}{\text{dy}}+\frac{\text{x}}{1+\text{y}^2}=\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}\ \dots(1)$
Clearly, it is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ where $\text{P}=\frac{1}{1+\text{y}^2}$ $\text{Q}=\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}$ $\therefore$ I.F. $=\text{e}^{\int\text{Pdy}}$ $=\text{e}^{\int\frac{1}{1+\text{y}^2}\text{dy}}$ $=\text{e}^{{\tan^{-1}\text{y}}}$ Multiplying both sides of (1) by $\text{e}^{\tan^{-1}\text{y}},$ we get $\text{e}^{\tan^{-1}\text{y}}\Big(\frac{\text{dx}}{\text{dy}}+\frac{\text{x}}{1+\text{y}^2}\Big)=\text{e}^{\tan^{-1}\text{y}}\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}$ $\Rightarrow\ \text{e}^{\tan^{-1}\text{y}}\frac{\text{dx}}{\text{dy}}+\frac{\text{x}\text{e}^{\tan^{-1}\text{x}}}{1+\text{y}^2}=\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}$ Integrating both sides with respect to y, we get $\text{x}\text{e}^{\tan^{-1}\text{y}}=\int\frac{\text{e}^{2\tan^{-1}\text{y}}}{1+\text{y}^2}\text{dy + C}$ $\Rightarrow\ \text{x}\text{e}^{\tan^{-1}\text{y}}=\text{I + C}\ \dots(2)$ Here, $\text{I}=\int\frac{\text{e}^{2\tan^{-1}\text{y}}}{1+\text{y}^2}\text{dy}$ Putting $\tan^{-1}\text{y = t},$ we get $\frac{1}{1+\text{y}^2}\text{dy = dt}$ $\therefore\ \text{I}=\int\text{e}^{2\text{t}}\text{dt}$ $=\frac{\text{e}^{2\text{t}}}{2}$ $=\frac{\text{e}^{2\tan^{-1}\text{y}}}{2}$ Putting the value of I in (2), we get $\text{x}\text{e}^{\tan^{-1}\text{y}}=\frac{\text{e}^{2\tan^{-1}\text{y}}}{2}+\text{C}$ $\Rightarrow\ 2\text{x}\text{e}^{\tan^{-1}\text{y}}=\text{e}^{2\tan^{-1}\text{y}}+2\text{C}$ $\Rightarrow\ 2\text{x}\text{e}^{\tan^{-1}\text{y}}=\text{e}^{2\tan^{-1}\text{y}}+\text{K}$ (where K = 2C) Hence, $2\text{x}\text{e}^{\tan^{-1}\text{y}}=\text{e}^{2\tan^{-1}\text{y}}+\text{K}$ is the required solution. View full question & answer→Question 984 Marks
Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.
AnswerLet F(x, y) be the curve passing through the origin. At point (x, y), the slope of the curve will be $\frac{\text{dy}}{\text{dx}}.$ According to the given infirmation: $\frac{\text{dy}}{\text{dx}}=\text{x}+\text{y}$ $\Rightarrow\frac{\text{dy}}{\text{dx}}-\text{y}=\text{x}$ This is a linear differential equation of the form: $\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q}\ (\text{where p}=-1\ \text{and}\ \text{Q}=\text{x})$ $\text{Now, I.F}=\text{e}^{\int\text{pdx}}=\text{e}^{\int(-1)\text{dx}}=\text{e}^{-\text{x}}.$ The general solution of the given differential equation is given by the relation, $\text{y(I.F)}=\int(\text{Q}\times\text{I.F.})\text{dx}+\text{C}$ $\Rightarrow\text{ye}^{-\text{x}}=\int\text{xe}^{-\text{x}}\text{dx}+\text{C}\ \ ...(1)$ $\text{Now},\int\text{xe}^{-\text{x}}\text{dx}=\text{x}\int\text{e}^{-\text{x}}\text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\text{x})\cdot\int\text{e}^{-\text{x}}\text{dx}\Big]\text{dx}.$ $=-\text{xe}^{-\text{x}}+\int-\text{e}^{-\text{x}}\text{dx}$ $=-\text{xe}^{-\text{x}}+(-\text{e}^{-\text{x}})$ $=-\text{e}^{-\text{x}}(\text{x}+1)$ Substituting in equation (1), we get: $\text{ye}^{\text{x}}=-\text{e}^{-\text{x}}(\text{x}+1)+\text{C}$ $\Rightarrow\text{y}=-(\text{x}+1)+\text{Ce}^{\text{x}}$ $\Rightarrow\text{x}+\text{y}+1=\text{Ce}^\text{x}\ \ ...(2)$ The curve passes through the origin. Therefore, equation (2) becomes:
$1=\text{C}$
$\Rightarrow\text{C}=1$
Substituting C = 1 in equation (2), we get: $\text{x}+\text{y}+1=\text{e}^\text{x}$ Hence, the required equation of curve passing through the origin is $\text{x}+\text{y}+1=\text{e}^\text{x}.$ View full question & answer→Question 994 Marks
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\cos^3\text{x}\sin^2\text{x}+\text{x}\sqrt{2\text{x}+1}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=\cos^3\text{x}\sin^2\text{x}+\text{x}\sqrt{2\text{x}+1}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(\cos^3\text{x}\sin^2\text{x}+\text{x}\sqrt{2\text{x}+1})\text{dx}$
Integrating both sides, we get
$\int\frac{\text{dy}}{\text{dx}}=\int(\cos^3\text{x}\sin^2\text{x}+\text{x}\sqrt{2\text{x}+1})\text{dx}$
$\Rightarrow\text{y}=\int\cos^3\text{x}\sin^2\text{x dx}+\int\text{x}\sqrt{2\text{x}+1}\text{dx}$
$\Rightarrow\text{y}=\text{I}_1+\text{I}_2\ ...(1)$
Where
$\text{I}_1=\int\cos^3\text{x}\sin^2\text{x dx}$
$\text{I}_2=\int\text{x}\sqrt{2\text{x}+1}\text{dx}$
Now,
$\text{I}_1=\int\cos^3\text{x}\sin^2\text{x dx}$
$=\int\sin^2\text{x}(1-\sin^2\text{x})\cos\text{x dx}$
Putting $\text{t}=\sin\text{x},$ we get
$\text{dt}=\cos\text{x dx}$
$\Rightarrow\text{I}_1=\int\text{t}^2(1-\text{t}^2)\text{dt}$
$=\int(\text{t}^2-\text{t}^4)\text{dt}$
$=\frac{\text{t}^3}{3}-\frac{\text{t}^5}{5}+\text{C}_1$
$=\frac{\sin^3\text{x}}{3}-\frac{\sin^5\text{x}}{5}+\text{C}_1$
$\text{I}_2=\int\text{x}\sqrt{2\text{x}+1}\text{dx}$
Putting $\text{t}^2=2\text{x}+1$ we get,
$2\text{t dt}=2\text{dx}$
$\Rightarrow\text{t dt}=\text{dx}$
Now,
$\text{I}_2=\int\Big(\frac{\text{t}^2-1}{2}\Big)\text{t}\times\text{t}\text{ dt}$
$=\frac{1}{2}\int(\text{t}^4-\text{t}^2)\text{dt}$
$=\frac{\text{t}^5}{10}-\frac{\text{t}^3}{6}+\text{C}_2$
$=\frac{(2\text{x}+1)\frac{5}{2}}{10}-\frac{(2\text{x}+1)^{\frac{3}{2}}}{6}+\text{C}_2$
Putting the value of I1 and I2 in (1), we get
$\text{y}=\frac{\sin^3\text{x}}{3}-\frac{\sin^5\text{x}}{5}+\text{C}_1+\frac{(2\text{x}+1)^{\frac{5}{2}}}{6}+\text{C}_2$
$\text{y}=\frac{\sin^3\text{x}}{3}-\frac{\sin^5\text{x}}{5}+\frac{(2\text{x}+1)^{\frac{5}{2}}}{10}-\frac{(2\text{x}+1)^{\frac{3}{2}}}{6}+\text{C}$
Hence, $\text{y}=\frac{\sin^3\text{x}}{3}-\frac{\sin^5\text{x}}{5}+\frac{(2\text{x}+1)^{\frac{5}{2}}}{10}-\frac{(2\text{x}+1)^{\frac{3}{2}}}{6}+\text{C}$ is the solution to the given differential equation.
View full question & answer→Question 1004 Marks
Find the equation of the curve through the point (1, 0) if the slope of the tangent to the curve at any point (x, y) is $\frac{\text{y}-1}{\text{x}^2+\text{x}}.$
AnswerIt is given that, slope of tangent to the curve at any point (x, y) is $\frac{\text{y}-1}{\text{x}^2+\text{x}}.$
$\therefore\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x},\text{y})}=\frac{\text{y}-1}{\text{x}^2+\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}-1}{\text{x}^2+\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{y}-1}=\frac{\text{dx}}{\text{x}^2+\text{x}}$
On integrating both sides, we get
$\int\frac{\text{dy}}{\text{y}-1}=\int\frac{\text{dx}}{\text{x}^2+\text{x}}$
$\Rightarrow\int\frac{\text{dy}}{\text{y}-1}=\int\frac{\text{dx}}{\text{x}(\text{x}+1)}$
$\Rightarrow\int\frac{\text{dy}}{\text{y}-1}=\int\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}+1}\Big)\text{dx}$
$\Rightarrow\log(\text{y}-1)=\log\text{x}-\log(\text{x+1})+\log\text{C}$
$\Rightarrow\log(\text{y}-1)=\log\Big(\frac{\text{x}\text{C}}{\text{x}+1}\Big)$
Since, the given curve passes through point (1, 0)
$\therefore0-1=\frac{1.\text{C}}{1+1}$
$\Rightarrow\text{C}=-2$
The particular solution is
$\text{y}-1=\frac{-2\text{x}}{\text{x}+1}$
$\Rightarrow(\text{y}-1)(\text{x}+1)=-2\text{x}$
$\Rightarrow(\text{y}-1)(\text{x}+1)+2\text{x}=0$
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