Question 12 Marks
$\text{Find} \frac{\text{dy}}{\text{dx}} \text{at } x = 1, \text{y} = \frac{\pi}{4} \text{if } { \sin}^{2}\text{y} + \cos x\text{y = K}.$
AnswerFrom the given equation
$2\sin \text{y } \cos \text{ y}. \frac{\text{dy}}{\text{dx}} - \sin \text{xy}.\bigg[\text{x}. \frac{\text{dy}}{\text{dx}} + \text{y . 1}\bigg] = 0$
$\Rightarrow \frac{\text{dy}}{\text{dx}} = \frac{\text{y} \sin \text{xy}}{\sin 2\text{y - x}\sin \text{(xy)}}$
$\therefore \frac{\text{dy}}{\text{dx}}\bigg|_{\text{x = 1, y =} \frac{\pi}{4}} = \frac{\pi}{4(\sqrt{2 - 1)}}$
View full question & answer→Question 22 Marks
Find the differential equation representing the family of curves $y = ae^{bx+5}$, where a and b are arbitrary constants.
Answer$\text{y} = \text{ae}^{\text{bx}+5}$ Take log on both sides$\log\text{y} = \log \big(\text{ae}^{\text{bx}+5}\big)$
$\log\text{y} = \log\text{a} + \log\big(\text{e}^{\text{bx}+5}\big)$ Differentiate both sides w.r.t. x $\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=0+\frac{\text{d}}{\text{dx}}(\text{bx}+5)$ $\frac{1}{\text{y}}\frac{\text{d}\text{y}}{\text{dx}}=\text{b}$ $\Rightarrow\frac{1}{\text{y}}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=0$ $\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{y}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=0$ i.e. the required differential $eq^n$.
View full question & answer→Question 32 Marks
Form the differential equation representing the family of curves $y = e^{2x}(a + bx)$, where ‘a’ and ‘b’ are arbitrary constants.
AnswerGiven: $y = e^{2x}(a + bx)$
Differentiating the above equation, we get
$\frac{\text{dy}}{\text{dx}}=\text{be}^{2\text{x}}+2(\text{a}+\text{bx})\text{e}^{2\text{x}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{be}^{2\text{x}}+2\text{y}\ \dots(\text{i})$ $\big[\therefore\text{y}=\text{e}^{2\text{x}}(\text{a}+\text{bx})\big]$
Differentiating the above equation, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{be}^{2\text{x}}+2\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\Big(\frac{\text{dy}}{\text{dx}}-2\text{y}\Big)+2\frac{\text{dy}}{\text{dx}}$ $\Big[\therefore\ $from (i) we get, $\text{be}^{2\text{x}}=\frac{\text{dy}}{\text{dx}}-2\text{y}\Big]$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=4\frac{\text{dy}}{\text{dx}}-4\text{y}$
Hence, the required differential equation is $\frac{\text{d}^2\text{y}}{\text{dx}^2}-4\frac{\text{dy}}{\text{dx}}+4\text{y}=0$
View full question & answer→Question 42 Marks
Determine the order and degree of the following differential equations. state also whether they are linear or non linear.
$\text{x}+\Big(\frac{\text{dy}}{\text{dx}}\Big)=\sqrt{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}$
Answer$\text{x}+\Big(\frac{\text{dy}}{\text{dx}}\Big)=\sqrt{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)=\Big(1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big)^{\frac{1}{2}}$
Squaring both sides, we get
$\Rightarrow\Big(\text{x}+\frac{\text{dy}}{\text{dx}}\Big)^2=1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
$\Rightarrow\text{x}^2+2\text{x}\frac{\text{dy}}{\text{dx}}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
$\Rightarrow2\text{x}\frac{\text{dy}}{\text{dx}}+\text{x}^2=1$
In this differential equation, the order of the highest order derivative is 1 and the power is 1. So, it is a differential equation of order 1 and degree 1.
Hence, it is a linear differential equation.
View full question & answer→Question 52 Marks
Form the differential equation from the following primitives where constants are arbitrart:$\text{y}^2=4\text{ax}$
AnswerThe equation of family of curves is
$\text{y}^2=4\text{ax}$
where a is an arbitrary constant. This equation contains only one arbitrary constant, so we shall get a differential equation of first order.
Differentiating equation (1) with respect to x, we get
$2\text{y}\frac{\text{dy}}{\text{dx}}=4\text{a}$
$\Rightarrow\frac{\text{y}}{2}\frac{\text{dy}}{\text{dx}}=\text{a}\ ...(2)$
Putting the value of a in equation (1), we get
$\text{y}^2=4\frac{\text{y}}{2}\frac{\text{dy}}{\text{dx}}\text{x}$
$\Rightarrow\text{y}=2\text{x}\frac{\text{dy}}{\text{dx}}$
It is the required differential equation.
View full question & answer→Question 62 Marks
Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=(\cos^2\text{x}-\sin^2\text{x})\cos^2\text{y}$
Answer$\frac{\text{dy}}{\text{dx}}=(\cos^2\text{x}-\sin^2\text{x})\cos^2\text{y}$
$\frac{\text{dy}}{\cos^2\text{y}}=(\cos^2\text{x}-\sin^2\text{x})\text{dx}$
$\int\sec^2\text{y dy}=\int\cos2\text{x dx}$
$\tan\text{y}=\frac{\sin2\text{x}}{2}+\text{C}$
View full question & answer→Question 72 Marks
If sinx is an integrating factor of the differential equation $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ than write the value of P.
AnswerIt is given that sinx is the intergrating factor of the differential equation $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}.$
$\text{e}^{\int\text{P}\text{dy}}=\sin\text{x}$
$\Rightarrow \int \text{P}\ \text{dx}=\log|\sin\text{x}|$
$\Rightarrow \int \text{P}\ \text{dx}=\int\cot\text{x}\ \text{dx}$
$\Rightarrow \text{P}=\cot\text{x}$
View full question & answer→Question 82 Marks
In each of the verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
$\text{y} = \text{cos} \ \text{x} + \text{C} \ :\ \text{y}' + \text{sin} \ \text{x} = 0$
AnswerGiven: y = cos x + C .....(i)
To prove: y is a solution of the differential equation y' + sin x = 0 .....(ii)
Proof: From eq. (i), y' = - sin x
$\therefore$ L.H.S. of eq. (ii), y' + sin x = - sin x + sin x = 0 = R.H.S.
Hence, y given by eq. (i) is a solution of y' + sin x = 0.
View full question & answer→Question 92 Marks
Determine the order and degree of the following differential equations. state also whether they are linear or non linear.
$\frac{\text{d}^4\text{y}}{\text{dx}^4}=\Big\{\text{c}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big\}^{\frac{3}{2}}$
Answer$\frac{\text{d}^4\text{y}}{\text{dx}^4}=\Big[\text{c}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]^{\frac{3}{2}}$
$\Rightarrow\Big(\frac{\text{d}^4\text{y}}{\text{dx}^4}\Big)^2=\Big[\text{c}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]^3$
$\Rightarrow\Big(\frac{\text{d}^2\text{y}}{\text{dx}^4}\Big)^2=\text{c}^3+\Big(\frac{\text{dy}}{\text{dx}}\Big)^6+3\text{c}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+3\text{c}^2\Big(\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow\Big(\frac{\text{d}^2\text{y}}{\text{dx}^4}\Big)^2-\Big(\frac{\text{dy}}{\text{dx}}\Big)^6-3\text{c}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-3\text{c}^2\Big(\frac{\text{dy}}{\text{dx}}\Big)-\text{c}^3=0$
The highest order differential coefficient is $\Big(\frac{\text{d}^2\text{y}}{\text{dx}^4}\Big)$ and its power is 2.
It is a non-linear differential equation with order 4 and degree 2.
View full question & answer→Question 102 Marks
In each of the form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b} }=1$
Answer$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b} }=1$
Differentiating both sides of the given equation with respect to x, we get:
$\frac{1}{\text{a}}+\frac{1}{\text{b}} \frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow \frac{1}{\text{a}}+\frac{1}{\text{b}}\text{y}'=0$
Again, differentiating both sides with respect to x, we get:
$0+\frac{1}{\text{b}}\text{y}''=0$
$\Rightarrow \frac{1}{\text{b}}\text{y}''=0$
$\Rightarrow \text{y}'' = 0$
Hence, the required differential equation of the given curve is y'' = 0.
View full question & answer→Question 112 Marks
Find the equation of the curve which passes through the point (3, -4) and has slope $\frac{2\text{y}}{\text{x}}$ at any point (x, y) on it.
AnswerAccording to the quation,
$\frac{\text{dy}}{\text{dx}}=\frac{2\text{y}}{\text{x}}$
$\Rightarrow \frac{1}{2\text{y}}\text{dy}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides with respect to x, we get
$\int\frac{1}{2\text{y}}\text{dy}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow \frac{1}{2}\log|\text{y}|=\log|\text{x}|+\text{C}$
Since the curve passes though (3, -4), it satisfies the above equation.
$\therefore \frac{1}{2}\log|-4|=\log|\text{x}|+\text{C}$
$\Rightarrow \log |2|-\log|3|+\text{C}$
$\Rightarrow \text{C}=\log|\frac{2}{3}|$
Putting the value of C, we get
$\log|\text{y}|=2\log|\text{x}|+2\log|\frac{2}{3}|$
$\log|\text{y}|=\log|\frac{4}{9}\text{x}^{2}|$
$\text{y}=\pm\frac{4}{9}\text{x}^{2}$
$9\text{y}-4\text{x}^{2}=0$
The given point does not satisfy the equation $9\text{y}-4\text{x}^{2}=0$
$\therefore 9\text{y}+4\text{x}^{2}=0$
View full question & answer→Question 122 Marks
Determine the order and degree of the following differential equations. state also whether they are linear or non linear.
$\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\frac{1}{\frac{\text{dy}}{\text{dx}}}=2$
Answer$\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\frac{1}{\Big(\frac{\text{dy}}{\text{dx}}\Big)}=2$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)^3+1=2\Big(\frac{\text{dy}}{\text{dx}}\Big)$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)^3-2\Big(\frac{\text{dy}}{\text{dx}}\Big)+1=0$
This is a polynomial in $\frac{\text{dy}}{\text{dx}}.$
The highest order differential coefficient is $\frac{\text{dy}}{\text{dx}}$ and its power is 3.
So, it is a non linear differential equation with order 1 and degree 3.
View full question & answer→Question 132 Marks
In each of the verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
$\text{y} = \text{Ax} \ :\ \text{xy}' = \text{y} (\text{x} \neq0)$
AnswerGiven: y = Ax .....(i)
To prove: y given by eq. (i) is a solution of differential equation $\text{xy}' = \text{y}(\text{x} \neq0)\ ....(\text{ii})$
Proof: From eq. (i) y' = A(1) = A
L.H.S. of eq. (ii) = xy' = xA = Ax = y = R.H.S. of eq. (ii)
$\therefore$ y given by eq. (i) is a solution of differential equation $\text{xy}' = \text{y}(\text{x} \neq 0).$
View full question & answer→Question 142 Marks
Form the differential equation from the following primitives where constants are arbitrart:$\text{y}=\text{ax}^2+\text{bx}+\text{c}$
AnswerThe equation of family of curves is
$\text{y}=\text{ax}^2+\text{bx}+\text{c}\ ...(1)$
where a, b and c is an arbitrary constant. so, we shall get a differential equation of third order.
Differentiating equation (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=2\text{ax}+\text{b}\ ...(2)$
Differentiating equation (2) with respect to x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}}=2\text{a}\ ...(3)$
Differentiating equation (3) with respect to x, we get
$\frac{\text{d}^3\text{y}}{\text{dx}3}=0$
It is the required differential equation.
View full question & answer→Question 152 Marks
Determine the order and degree of the following differential equations. state also whether they are linear or non linear.
$\text{s}^2\frac{\text{d}^2\text{t}}{\text{ds}^2}+\text{st}\frac{\text{dt}}{\text{ds}}=\text{s}$
Answer$\text{s}^2\frac{\text{d}^2\text{t}}{\text{ds}^2}+\text{st}\frac{\text{dt}}{\text{ds}}=\text{s}$
$\Rightarrow\text{s}\frac{\text{d}^2\text{t}}{\text{ds}}^2+\text{t}\frac{\text{dt}}{\text{ds}}=1$
In this differential equation, the order of the highest order derivative is 2 and its power is 1. So, it is a differential equation of order 2 and degree 1.
It is a non-linear differential equation, as it contains the product of the dependent variable (t) and its differential co-efficient $\Big(\frac{\text{dt}}{\text{ds}}\Big).$
View full question & answer→Question 162 Marks
Determine the order and degree of the following differential equations. state also whether they are linear or non linear.
$\frac{\text{d}^2\text{y}}{\text{dx}^2}+4\text{y}=0$
Answer$\frac{\text{d}^2\text{y}}{\text{dx}^2}+4\text{y}=0$
It is a linear differential equation.
The highest order differential coefficient is $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ and its power is 1.
So, It is a linear differential equation with order 2 and degree 1.
View full question & answer→Question 172 Marks
For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
| $\text{y}=\text{a e}^{\text{x}}+\text{b e}^{-\text{x}}+\text{x}^2$ |
: |
$\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}+2\frac{\text{dy}}{\text{dx}}-\text{xy}+\text{x}^2-2=0$ |
Answer$\text{Here,}\ \ \text{y}=\text{a e}^{\text{x}}+\text{b e}^{-\text{x}}+\text{x}^2$
$\therefore\ \ \frac{\text{dy}}{\text{dx}}=\text{a e}^{\text{x}}-\text{b e}^{-\text{x}}+2\text{x}$
$\therefore\ \ \frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{a e}^{\text{x}}+\text{b e}^{-\text{x}}+2$
$\therefore\ \ \text{L.H.S.}=\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{y}+\text{x}^2-2$
$=\text{a e}^{\text{x}}+\text{b e}^{-\text{x}}+2-(\text{a e}^{\text{x}}+\text{b e}^{-\text{x}}+\text{x}^2)+\text{x}^2-2$
$=\text{a e}^{\text{x}}+\text{b e}^{-\text{x}}+2-\text{a e}^{\text{x}}+\text{b e}^{-\text{x}}+\text{x}^2+\text{x}^2-2$
$=0=\text{R.H.S.}$
$\therefore\ \ \text{y}=\text{a e}^{\text{x}}+\text{b e}^{-\text{x}}+\text{x}^2$
is a solution of the given differential equation.
View full question & answer→Question 182 Marks
Determine the order and degree of the following differential equations. state also whether they are linear or non linear.
$\sqrt{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}=\Big(\text{c}\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)^{\frac{1}{3}}$
Answer$\sqrt{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}=\Big(\text{c}\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)^{\frac{1}{3}}$
Squaring both sides, we get
$\Rightarrow1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\Big(\text{c}\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)^{\frac{2}{3}}$
Taking cubes of both sides, we get
$\Rightarrow\Big(\text{c}\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)=\Big[1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]^3$
$\Rightarrow\Big(\text{c}\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)=1+3\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+3\Big(\frac{\text{dy}}{\text{dx}}\Big)^4+\Big(\frac{\text{dy}}{\text{dx}}\Big)^6$
In this differential equation, the order of the highest order derivative is 2 and its power is 2. So, it is a differential equation of order 2 and degree 2.
It is a non-linear differential equation, as its degree is more than 1.
View full question & answer→Question 192 Marks
Solve the following differential equations:
$(\text{y}^2+1)\text{dx}-(\text{x}^2+1)\text{dy}=0$
Answer$(\text{y}^2+1)\text{dx}-(\text{x}^2+1)\text{dy}=0$
$(\text{y}^2+1)\text{dx}=(\text{x}^2+1)\text{dy}$
$\int\frac{\text{dy}}{\text{y}^2+1}=\int\frac{\text{dx}}{\text{x}^2+1}$
$\tan^{-1}\text{y}=\tan^{-1}\text{x + C}$
View full question & answer→Question 202 Marks
Determine the order and degree of the following differential equations. state also whether they are linear or non linear.
$\frac{\text{d}^3\text{x}}{\text{dt}^3}+\frac{\text{d}^3\text{x}}{\text{dt}^2}+\Big(\frac{\text{ dx}}{\text{dt}}\Big)^2=\text{e}^\text{t}$
AnswerIn this differential equation, the order of the highest order derivative is 3 and its power is 1. So, it is a differential equation of order 3 and degree 1.
It is a non-linear differential equation because the differential coefficient $\frac{\text{dx}}{\text{dt}}$ has exponent 2, which is greater than 1.
View full question & answer→Question 212 Marks
Determine the order and degree of the following differential equations. state also whether they are linear or non linear.
$5\frac{\text{d}^2\text{y}}{\text{dx}^2}=\Big\{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big\}^{\frac{3}{2}}$
Answer$5\frac{\text{d}^2\text{y}}{\text{dx}^2}=\Big\{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big\}^{\frac{3}{2}}$
$\Big\{5\Big(\frac{\text{d}^2\text{y}}{\text{dx}}\Big)^2\Big\}=\Big\{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big\}^3$
$25\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)^2=1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^6+3\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+3\Big(\frac{\text{dy}}{\text{dx}}\Big)^4$
$25\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)^2-\Big(\frac{\text{dy}}{\text{dx}}\Big)^6-3\Big(\frac{\text{dy}}{\text{dx}}\Big)^4-3\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-1=0$
The highest order differential coefficient is $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ and its power is 2.
So, it is a non linear differential equation with order 2 and degee 2.
View full question & answer→Question 222 Marks
Form the differential equation from the following primitives where constants are arbitrart:$\text{xy}=\text{a}^2$
AnswerThe equation of family of curves is
$\text{xy}=\text{a}^2\ ...(1)$
where a is an arbitrary constant.
This equation contains only one arbitrary constant, so we shall get a differential equation of first order.
Differentiating equation (1) with respect to x, we get
$\text{y}+\text{x}\frac{\text{dy}}{\text{dx}}=0$
It is the required differential equation.
View full question & answer→Question 232 Marks
Define a differential equation.
AnswerAn equation containing an independent variable, a dependent variable and differential cofficients of the dependent variable with reapect to the independent variable is called a differential equation.
for example: $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}$
View full question & answer→Question 242 Marks
Form the differential equation from the following primitives where constants are arbitrart:$\text{y}=\text{cx}+2\text{c}^2+\text{c}$
AnswerThe equation of family of curves is
$\text{y}=\text{cx}+2\text{c}^2+\text{c}\ ...(1)$
where c is an arbitrary constant.
This equation contains only one arbitrary constant, so we shall get a differential equation of first order.
Differentiating equation (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\text{c}\ ...(2)$
$\Rightarrow\frac{\text{y}}{2}\frac{\text{dy}}{\text{dx}}=\text{a}\ ...(2)$
Putting the value of a in equation (1), we get
$\text{y}=\text{x}\frac{\text{dy}}{\text{dx}}+2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\Big(\frac{\text{dy}}{\text{dx}}\Big)^3$
It is the required differential equation.
View full question & answer→Question 252 Marks
In each of the verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
$\text{x + y} = \text{tan}^{–1}\ \text{y} \ \ :\ \text{y}^2 \text{y}' + \text{y}^2 + 1 = 0$
AnswerGiven: $x + y = \tan^{-1} y .....(i)$
To prove: y given by eq. (i) is a solution of differential equation $y^2y' + y^2 + 1 = 0 ....(ii)$
Proof: Differentiating both sides of eq. (i) w.r.t x, we have
$1+\text{y}'=\frac{1}{1+\text{y}^2}\text{y}' \ \Rightarrow \ \ (1+\text{y}')(1+\text{y}^2 ) = \text{y}'$
$\Rightarrow\ \ \ \ 1+ \text{y}^2 + \text{y}'+ \text{y}' \text{y}^2 = \text{y}' $ $\Rightarrow\ \text{y}^2 \text{y}'+ \text{y}^2 +1 -0 = \text{eq}. (\text{ii})$
Hence, Function given by eq. (i) is a solution of $y^2 y' + y^2 + 1 = 0$
View full question & answer→Question 262 Marks
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}+\text{e}^\text{y}\text{x}^3$
Answer$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}+\text{e}^\text{y}\text{x}^3$
$\frac{\text{dy}}{\text{dx}}=\text{e}^\text{y}(\text{e}^\text{x}+\text{x}^3)$
$\int\text{e}^{-\text{y}}\text{dy}=\int(\text{e}^\text{x}+\text{x}^3)\text{dx}$
$-\text{e}^{-\text{y}}-\text{e}^\text{x}+\frac{\text{x}^4}{4}+\text{C}_1$
$\text{e}^\text{x}+\frac{\text{x}^4}{4}+\text{e}^{-\text{y}}=\text{C}$
View full question & answer→Question 272 Marks
Write the order of the differential equation $\text{y}=\text{x}\frac{\text{dy}}{\text{dx}}+\text{a}\sqrt{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{2}}.$
AnswerWe have,
$\text{y}=\text{x}\frac{\text{dy}}{\text{dx}}+\text{a}\sqrt{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{2}}$
$\text{y}-\text{x}\frac{\text{dy}}{\text{dx}}=\text{a}\sqrt{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{2}}$
Squaring both sides, we get
$\Big(\text{y}-\text{x}\frac{\text{dy}}{\text{dx}}\Big)^{2}=\left\{\text{a}\sqrt{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{2}}\right\}^{2}$
$\text{y}^{2}-\text{x}^{2}\Big(\frac{\text{dy}}{\text{dx}}\Big)^{2}-2\text{xy}\frac{\text{dy}}{\text{dx}}=\text{a}^{2}\left\{{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{2}}\right\}^{2}$
$\text{y}^{2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{2}(\text{x}^{2}-\text{a}^{2})-2\text{xy}\frac{\text{dy}}{\text{dx}}=\text{a}^{2}$
From the above equation, we see that the highest order is 1.
So, its order is 1 and the power of the order is 2.
Thus, it is differential equation of order 1 and degree 2.
View full question & answer→Question 282 Marks
Determine the order and degree of the following differential equations. state also whether they are linear or non linear.
$\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\text{xy}=0$
Answer$\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\text{xy}=0$
In this differential equation, the order of the highest order derivative is 2 and its power is 1. So, it is a differential equation of order 2 and degree 1.
It is a non-linear differential equation, as the differential coefficient $\frac{\text{dy}}{\text{dx}}$ has exponent 2, which is greater than 1.
View full question & answer→Question 292 Marks
For the following differntial equations verify that the accompanying function is a solution:
| Differential equation |
Function |
| $\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}$ |
$\text{y}=\text{ax}$ |
AnswerWe have
$\text{y}=\text{ax}\ ...(1)$
Given differential equation
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\text{a}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}={\text{y}}$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 302 Marks
Determine the order and degree of the following differential equations. state also whether they are linear or non linear.
${3}\sqrt{\frac{\text{d}^2\text{y}}{\text{dx}^2}}=\sqrt{\frac{\text{dy}}{\text{dx}}}$
Answer${3}\sqrt{\frac{\text{d}^2\text{y}}{\text{dx}^2}}=\sqrt{\frac{\text{dy}}{\text{dx}}}$
$\Rightarrow\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)^{\frac{1}{3}}=\Big(\frac{\text{dy}}{\text{dx}}\Big)^\frac{1}{2}$
Taking cubes of both the sides, we get
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\Big(\frac{\text{dy}}{\text{dx}}\Big)^{\frac{3}{2}}$
Squaring both the sides, we get
$\Rightarrow\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)^2=\Big(\frac{\text{dy}}{\text{dx}}\Big)^3$
$\Rightarrow\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)^2-\Big(\frac{\text{dy}}{\text{dx}}\Big)^3=0$
In this differential equation, the order of the highest order derivative is 2 and its power is 2. So, it is a differential equation of order 2 and degree 2.
Thus, it is a non-linear differential equation, as its degree is 2, which is greater than 1.
View full question & answer→Question 312 Marks
In each of the verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
$\text{y} = \text{x}^2 + 2\text{x} +\text{C}\ :\ \text{y}' – 2\text{x} – 2 = 0$
AnswerGiven: $y = x^2 + 2x + C .....(i)$
To prove: y is a solution of the differential equation $y' - 2x - 2 = 0 .....(ii)$
Proof: From eq. $y' = 2x + 2$
$\therefore$ L.H.S. of eq. $(ii), y' - 2x - 2 = (2x + 2) - 2x - 2$
$= 2x + 2 - 2x - 2 = 0 =$ R.H.S.
Hence, $y$ given by eq. $(i)$ is a solution of $y' - 2x - 2 = 0.$
View full question & answer→Question 322 Marks
Show that $\text{y}=\text{e}^{-\text{x}}+\text{ax}+\text{b}$ is solution of the differential equation $\text{e}^\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}=1$
AnswerWe have
$\text{y}=\text{e}^{-\text{x}}+\text{ax}+\text{b}\ ...(1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=-\text{e}^{-\text{x}}+\text{a}\ ...(2)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{e}^{-\text{x}}$
$\Rightarrow\text{e}^\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}=1$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 332 Marks
Determine the order and degree of the following differential equations. state also whether they are linear or non linear.
$\text{y}\frac{\text{d}^2\text{x}}{\text{dy}^2}=\text{y}^2+1$
Answer$\text{y}\frac{\text{d}^2\text{x}}{\text{dy}^2}=\text{y}^2+1$ $\frac{\text{d}^2\text{x}}{\text{dy}^2}-\text{y}-\frac{1}{\text{y}}=0$The differential coefficient is $\frac{\text{d}^2\text{x}}{\text{dy}^2}$ and its power is 1.
So, it is a linear differential equation with order 2 and degree 1.
View full question & answer→Question 342 Marks
Determine the order and degree of the following differential equations. state also whether they are linear or non linear.
$2\frac{\text{d}^2\text{y}}{\text{dx}^2}+3\sqrt{1-\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}}=0$
Answer$2\frac{\text{d}^2\text{y}}{\text{dx}^2}+3\sqrt{1-\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}}=0$
$2\frac{\text{d}^2\text{y}}{\text{dx}^2}+3\sqrt{1-\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}}$
Squaring both the sides,
$4\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)^2=9\Big(1-\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}\Big)$
$4\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)^2+9\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+9\text{y}-9=0$
The highest order differential coefficient is $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ and its power is 2.
So, it is a non linear differential equation with order 2 and degee 2.
View full question & answer→Question 352 Marks
For each of the differential equations given in find the general solution:
$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=\text{x}^2$
AnswerThe given differential equation is:$\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q }{(\text{where p}}=\frac{1}{\text{x}}\text{and Q}=\text{x}^2\big)$
$\text{Now, I.F}=\text{e}^{\int\text{pdx}}=\text{e}^{\int\frac{1}{\text{x}}\text{dx}}=\text{e}^{\log\text{x}}=\text{x}.$
The solution of the given differential equation is given by the relation,
$\text{y}(\text{I.F})=\int(\text{Q}\times\text{I.F.})\text{dx}+\text{C}$
$\Rightarrow\ \text{y}\ (\text{x})=\int(\text{x}^2\cdot\text{x})\text{dx}+\text{C}$
$\Rightarrow\ \text{xy}=\int\text{x}^3\text{dx}+\text{C}$
$\Rightarrow\ \text{xy}=\frac{\text{x}^4}{4}+\text{C}$
This is the required general solution of the given differential equation.
View full question & answer→