Question 513 Marks
Find a particular solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=4\text{x}\ \text{cosec}\ \text{x}\ (\text{x}\neq0),$ $\text{given that y}=0\ \text{when x}=\frac{\pi}{2}.$
AnswerGiven: Differential equation $\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=4\text{x}\ \text{cosec}\ \text{x}$
Comparing this equation with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},\ \ \text{P}=\cot\text{x}\ \text{and}\ \text{Q}=4\text{x}\ \text{cosec}\ \text{x}$
$\int\text{P}\ \text{dx}=\int\cot\text{x}\ \text{dx}=\log\sin\text{x}$ $\text{I.F}=\text{e}^{\int\text{p dx}}=\text{e}^{\log\sin\text{x}}=\sin\text{x}$
The general solution is $\text{y}(\text{I.F})=\int\text{Q}(\text{I.F}.)\ \text{dx}+\text{c}$
$\Rightarrow\ \ \text{y}(\sin\text{x})=\int4\text{x}\ \text{cosec}\ \text{x}\ \sin\text{x}\ \text{dx}+\text{c}$ $\Rightarrow\ \ \text{y}(\sin\text{x})=4\int\text{x}.\frac{1}{\sin\text{x}}\sin\text{x}\ \text{dx}+\text{c}$
$\Rightarrow\ \ \text{y}(\sin\text{x})=4\int\text{x}\ \text{dx}+\text{c}=4\frac{\text{x}^2}{2}+\text{c}$ $\Rightarrow\ \ \text{y}\sin\text{x}=2\text{x}^2+\text{c}\ \ ...{(\text{i})}$
$\text{Now Putting y}=0,\text{x}=\frac{\pi}{2}\ \text{in eq. (i)},$ $0=2.\frac{\pi^{2}}{4}+\text{c}\ \ \Rightarrow\ \ \text{c}=\frac{-\pi^{2}}{2}$
$\text{Putting c}=\frac{-\pi^2}{2}\ \text{in eq. (i)},\ \ \text{y}\sin\text{x}=2\text{x}^2-\frac{\pi^2}{2}$
View full question & answer→Question 523 Marks
Solve the following differential equations:
$\sqrt{1+\text{x}^2}\ \text{dy}+\sqrt{1+\text{y}^2}\ \text{dx}=0$
Answer$\sqrt{1+\text{x}^2}\ \text{dy}+\sqrt{1+\text{y}^2}\ \text{dx}=0$
$\sqrt{1+\text{x}^2}\ \text{dy}=-\sqrt{1+\text{y}^2}\ \text{dx}$
$\int\frac{\text{dy}}{\sqrt{1+\text{y}^2}}=-\int\frac{\text{dx}}{\sqrt{1+\text{x}^2}}$
$\log|\text{y}+\sqrt{1+\text{y}^2}|=-\log|\text{x}+\sqrt{1+\text{x}^2}|=\log|\text{c}|$
$(\text{y}+\sqrt{1+\text{y}^2})(\text{x}+\sqrt{1+\text{x}^2})=\text{c}$
View full question & answer→Question 533 Marks
Solve the following equation
$\text{xy dy}=(\text{y}-1)(\text{x}+1)\text{dx}$
Answer$\text{xy dy}=(\text{y}-1)(\text{x}+1)\text{dx}$
$\frac{\text{y}}{\text{y}-1}\text{dy}=\frac{\text{x}+1}{\text{x}}\ \text{dx}$
$\int\Big(1+\frac{1}{\text{y}-1}\Big)\text{dy}=\int\Big(1+\frac{1}{\text{x}}\Big)\text{dx}$
$\text{y}+\log|\text{y}-1|=\text{x}+\log|\text{x}|+\text{C}$
$\text{y}-\text{x}=\log|\text{x}|-\log|\text{y}-1|+\text{C}$
View full question & answer→Question 543 Marks
Prove that $\text{x}^2-\text{y}^2=\text{c}(\text{x}^2+\text{y}^2)^2$ is the general solution of differential equation $(\text{x}^3-3\text{x y}^2) \text{dx}=(\text{y}^3-3\text{x}^2\text{y})\text{dy, where c}$ is a parameter.
Answer$\text{Here},\ \ \text{x}^2-\text{y}^2=\text{c}(\text{x}^2+\text{y}^2)^2\ \ ...{(1)}$
Differentiating w.r.t.x, we get,
$2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}=2\text{c}(\text{x}^2+\text{y}^2).\Big[2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big]$
$\text{x}-\text{y}\frac{\text{dy}}{\text{dx}}=2\text{c}(\text{x}^2+\text{y}^2).\Big[\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}\Big]\ \ ...(2)$
Dividing (2) by (1), we get.
$\frac{\text{x}-\text{y}\frac{\text{dy}}{\text{dx}}}{\text{x}^2-\text{y}^2}=\frac{2\Big(\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}\Big)}{\text{x}^2+\text{y}^2}$
$\therefore\ \ \text{x}(\text{x}^2+\text{y}^2)-\text{y}(\text{x}^2+\text{y}^2)\frac{\text{dy}}{\text{dx}}$ $=2\text{x}(\text{x}^2-\text{y}^2)+2\text{y}(\text{x}^2-\text{y}^2)\frac{\text{dy}}{\text{dx}}$
$\therefore\ \ \big[2\text{y}(\text{x}^2-\text{y}^2)+\text{y}(\text{x}^2+\text{y}^2)\big]\frac{\text{dy}}{\text{dx}}$ $=\text{x}(\text{x}^2+\text{y}^2)-2\text{x}(\text{x}^2-\text{y}^2)$
$\therefore\ \ (2\text{x}^2\text{y}-2\text{y}^3+\text{x}^2\text{y}+\text{y}^3)\frac{\text{dy}}{\text{dx}}$ $=\text{x}^3+\text{xy}^2-2\text{x}^3+2\text{xy}^2$
$\therefore\ \ (3\text{x}^2\text{y}-\text{y}^3)\frac{\text{dy}}{\text{dx}}=3\text{xy}^2-\text{x}^3$
$\therefore\ \ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}^3-3\text{xy}^2}{\text{y}^3-3\text{x}^2\text{y}}$ which is the required equation.
Hence the result.
View full question & answer→Question 553 Marks
Solve the following differential equations:
$\text{dy}+(\text{x}+1)(\text{y}+1)\text{dx}=0$
Answer$\text{dy}+(\text{x}+1)(\text{y}+1)\text{dx}=0$
$\text{dy}=-(\text{x}+1)(\text{y}+1)\text{dx}$
$\int\frac{\text{dy}}{\text{y}+1}=-\int(\text{x}+1)\text{dx}$
$\log|\text{y}+1|=-\frac{\text{x}^2}{2}-\text{x}+\text{C}$
$\log|\text{y}+1|+\frac{\text{x}^2}{2}+\text{x = C}$
View full question & answer→Question 563 Marks
Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x+y}}+\text{e}^{-\text{x+y}}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x+y}}+\text{e}^{-\text{x+y}}$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x+y}}+\text{e}^{-\text{x}+\text{y}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{y}}(\text{e}^{\text{x}}+\text{e}^{-\text{x}})$
$\Rightarrow\text{e}^{-\text{y}}\text{dy}=(\text{e}^{\text{x}}+\text{e}^{-\text{x}})\text{dx}$
Integrating both sides, we get
$\int\text{e}^{-\text{y}}\text{dy}=\int(\text{e}^{\text{x}}+\text{e}^{-\text{x}})\text{dx}$
$\Rightarrow-\text{e}^{-\text{y}}=\text{e}^{\text{x}}-\text{e}^{-\text{x}}+\text{C}$
$\Rightarrow\text{e}^{-\text{x}}-\text{e}^{-\text{y}}=\text{e}^{\text{x}}+\text{C}$
Hence, $\text{e}^{-\text{x}}-\text{e}^{-\text{y}}=\text{e}^{\text{x}}+\text{C}$ is the required solution.
View full question & answer→Question 573 Marks
In each of the form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.
y = e2x (a + bx)
Answery = e2x (a + bx) ....(1) Differentiating both sides with respect to x, we get: $\text{y}'=2\text{e}^{2\text{x}}(\text{a+bx}) + \text{e}^{2\text{x}}\cdot\text{b}$ $\Rightarrow \text{y}'=\text{e}^{2\text{x}} (2\text{a}+2\text{bx}+\text{b}) \ ....(2)$ MUltiplying equation (1) by (2) and than subtracting it equation (2), we get: $\text{y}'-2\text{y = e}^{2\text{x}} (2\text{a + 2bx + b)}-\text{e}^{2\text{x}}(2\text{a + 2bx})$
$\Rightarrow \text{y}'-2\text{y}=\text{be}^{2\text{x}} \ ...(3)$
Differentiating both sides with respect to x, we get: $\text{y}''-2\text{y}'=2\text{be}^{2\text{x}} \ ...(4)$ Dividing equation (4) by equation (3), we get: $\frac{\text{y}''-2\text{y}'}{\text{y}'-2\text{y}}=2$ $\Rightarrow \text{y}''-2\text{y}'=2\text{y}'-4\text{y}$ $\Rightarrow \text{y}''-4\text{y}'+4\text{y}=0$ This is the required differential equation of the given curve. View full question & answer→Question 583 Marks
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\sin^2\text{y}$
AnswerWe have $\frac{\text{dy}}{\text{dx}}=\sin^2\text{y}$ $\Rightarrow\text{dx}=\frac{1}{\sin^2\text{y}}$ $\Rightarrow\text{dx}=\text{cosec}^2\text{y dy}$ Integrating both sides, we get $\int\text{dx}=\text{cosec}^2\text{y dy}$ $\Rightarrow\text{x}=-\cot\text{y}+\text{C}$ $\Rightarrow\text{x}+\cot\text{y}=\text{C}$
Hence, $\Rightarrow\text{x}+\cot\text{y}=\text{C}$ is the required solution. View full question & answer→Question 593 Marks
For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
| $\text{y}=\text{e}^{\text{x}}(\text{a}\cos\text{x}+\text{b}\sin\text{x})$ | : | $\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\frac{\text{dy}}{\text{dx}}+2\text{y}=0$ |
AnswerThe given differential equation is
$\text{y}=\text{e}^{\text{x}}(\text{a}\cos\text{x}+\text{b}\sin\text{x})$
$\Rightarrow\ \text{e}^{-\text{x}}\text{y}=\text{a}\cos\text{x}+\text{b}\sin\text{x}\ \ ...(1)$
Differentiating (1) twice w.r.t. .x. we get
$\text{e}^{-\text{x}}\frac{\text{dy}}{\text{dx}}+\text{y e}^{-\text{x}}(-1)=-\text{a}\sin\text{x}+\text{b}\cos\text{x}$
$\text{and}\ \ \text{e}^{-\text{x}}\ \frac{\text{d}^2\text{y}}{\text{dx}^2}+\frac{\text{dy}}{\text{dx}}\text{e}^{-\text{x}}(-1)-\Big\{-\text{y e}^{-\text{x}}+\text{e}^{-\text{x}}\ \frac{\text{dy}}{\text{dx}}\Big\}$ $=-\text{a}\cos\text{x}-\text{b}\sin\text{x}$
$ \text{or}\ \ \text{e}^{-\text{x}}\ \frac{\text{d}^2\text{y}}{\text{dx}^2}-2\text{e}^{-\text{x}}\ \frac{\text{dy}}{\text{dx}}+\text{y e}^{-\text{x}}=-\text{y x}^{-\text{x}}\ \ [\because \text{of}\ (1)]$
$ \text{or}\ \ \text{e}^{-\text{x}}\ \frac{\text{d}^2\text{y}}{\text{dx}^2}-2\text{e}^{-\text{x}}\ \frac{\text{dy}}{\text{dx}}+2\text{y e}^{-\text{x}}=0$ $\ \text{or}\ \text{e}^{-\text{x}}\Big\{\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\ \frac{\text{dy}}{\text{dx}}+2\text{y}\Big\}=0$
$\ \text{or}\ \ \frac{\text{d}^2\text{y}}{\text{dx}^2}-2\ \frac{\text{dy}}{\text{dx}}+2\text{y}=0$
Hence the result.
View full question & answer→Question 603 Marks
In each of the verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
$\text{xy} = \text{log}\ \text{y} + \text{C} \ \ : \ \text{y}'=\frac{\text{y}^2}{1-\text{xy}}\ (\text{xy} \neq1)$
AnswerGiven: xy = log y + C ....(i) To prove: y given by eq. (i) is a solution of differential equation $\text{y}'=\frac{\text{y}^2}{1-\text{xy}} \ ...(\text{ii})$ Proof: Differentiating both sides of eq. (i) w.r.t x, we have $\text{xy}'+\text{y}(1)=\frac{1}{\text{y}}\text{y}'+0 \ $ $\Rightarrow \ \text{xy}'-\frac{\text{y}'}{\text{y}}=-\text{y} $ $ \Rightarrow \ \text{y}'\bigg(\text{x}-\frac{\text{1}}{\text{y}}\bigg)=-\text{y}$ $\Rightarrow \ \text{y}'\bigg(\frac{\text{xy}-1}{\text{y}}\bigg)=-\text{y}\ $ $\Rightarrow \ \text{y}'(\text{xy}-1)=-\text{y}^2$ $ \Rightarrow \ \text{y}'=\frac{-\text{y}^2}{\text{xy}-1}$ $\Rightarrow \ \ \text{y}'=\frac{-\text{y}^2}{-(1-\text{xy})}=\frac{\text{y}^2}{1-\text{xy}}$
Hence, Function (implicit) given by eq. (i) is a solution of ${\text{y}}'=\frac{\text{y}^2}{1-\text{xy}}.$ View full question & answer→Question 613 Marks
For each of the differential equations in find the general solution:
$\text{x}^5\frac{\text{dy}}{\text{dx}}=-\text{y}^5$
AnswerThe given differential equation is
$\text{x}^5\frac{\text{dy}}{\text{dx}}=-\text{y}^5 \ \ \text{or} \ \ \frac{\text{dy}}{\text{dx}}=-\frac{\text{y}^5}{\text{x}^5}$
Separating the variables, we get,
$\frac{1}{\text{y}^5}\text{dy}=-\frac{1}{\text{x}^5}\text{dx}$
Integrating, $\int \frac{\text{1}}{\text{y}^5}\text{dy}=-\int \frac{1}{\text{x}^5}\ \text{dx} \ \text{or}$ $ \int \text{y}^{-5}\text{dy}=-\int\text{x}^{-5}\text{dx}$
$\therefore\ \ \frac{\text{y}^{-4}}{-4}=-\frac{\text{x}^{-4}}{-4}+\text{c}'\ \text{or}$ $\frac{1}{-4\text{y}^4}=\frac{1}{4\text{x}^4}+\text{c}'$
$\text{or} \ \frac{1}{\text{x}^4}+\frac{1}{\text{y}^4}=-4\text{c}'$
$\text{or}\ \text{x}^{-4}+\text{y}^{-4}=\text{c}, \ \text{where c}=-4\text{c}'.$
View full question & answer→Question 623 Marks
For each of the differential equations in find the general solution:
$\frac{\text{dy}}{\text{dx}}=\text{sin}^{-1}\text{x}$
AnswerThe given differential equation is $\frac{\text{dy}}{\text{dx}}=\sin^{-1}\text{x}$
Separating the variables and integrating, $\int \text{dy}=\int\text{sin}^{-1}\text{x dx}$
$\therefore\ \int \text{1 dy}= \int \sin^{-1}\text{x}\cdot 1\ \text{dx}$ $\therefore \ \text{y}=(\text{sin}^{-1}\text{x}).\text{x}-\int\frac{1}{\sqrt{1-\text{x}^2}}.\text{x dx}$ $\text{or}\ \text{y = x sin}^{-1}\text{x}+\frac{1}{2}\int(1-\text{x}^2)^{-\frac{1}{2}}(-2\ \text{x}) \ \text{dx}$ $\therefore\ \text{y}=\text{x sin}^{-1}\text{x}+\frac{1}{2}\frac{(1-\text{x}^2)\frac{1}{2}}{\frac{1}{2}}+\text{c}$ $\therefore\ \text{y}= \text{x sin}^{-1}\text{x}+\sqrt{1-\text{x}^2}+\text{c}$ which is the required solution. View full question & answer→Question 633 Marks
Solve the differential equation $\frac{\text{dy}}{\text{dx}}+1=\text{e}^{\text{x+y}}.$
AnswerWe have $\frac{\text{dy}}{\text{dx}}+1=\text{e}^{\text{x+y}}\ ....(\text{i})$
Take $\text{x}+\text{y}=\text{t}$
$\Rightarrow1+\frac{\text{dy}}{\text{dx}}=\frac{\text{dt}}{\text{dx}}$
Substituting $\text{x}+\text{y}=\text{t}$ in equation (i) we get,
$\frac{\text{dt}}{\text{dx}}=\text{e}^\text{t}$
$\Rightarrow\text{e}^{-\text{t}\text{dt}}=\text{dx}$
$\Rightarrow-\text{e}^{-\text{t}}=\text{x}+\text{C}$
$\Rightarrow\frac{-1}{\text{e}^\text{x+y}}=\text{x}+\text{C}$
$\Rightarrow-1=(\text{x}+\text{C})\text{e}^\text{x+y}$
$\Rightarrow(\text{x}+\text{C})\text{ e}^\text{x+y}1=0$
View full question & answer→Question 643 Marks
Verify that $\text{y}=4\sin3\text{x}$ is a solution of the differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}+9\text{y}=0.$
AnswerWe have,
$\text{y}=4\sin3\text{x}\ ...(1)$
Differentiating both sides of equation (1) with respect to 3, we get
$\frac{\text{dy}}{\text{dx}}=12\cos3\text{x}\ ...(2)$
Differentiating both sides of equation (2) with respect to 3, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-36\sin3\text{x}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-9(4\sin3\text{x})$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-9\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}+9\text{y}=0$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 653 Marks
Solve the following differential equation
$\text{x}\frac{\text{dy}}{\text{dx}}+\cot\text{y}=0$
AnswerWe have, $\text{x}\frac{\text{dy}}{\text{dx}}+\cot\text{y}=0$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}-\cot\text{y}$
$\Rightarrow\frac{1}{\text{x}}\ \text{dx}=-\frac{1}{\cot\text{y}}\ \text{dy}$
$\Rightarrow\frac{1}{\text{x}}\ \text{dx}=-\tan\text{y dy}$
Integrating both sides, we get
$\int\frac{1}{\text{x}}\ \text{dx}=-\int\tan\text{y dy}$
$\Rightarrow\int|\text{x}|=-\int|\sec\text{y}|+\int\text{C}$
$\Rightarrow\int|\text{x}|=-\int|\cos\text{y}|+\int\text{C}$
$\Rightarrow\text{x}=\text{C}\cos\text{y}$
Hence, $\text{x}=\text{C}\cos\text{y}$ is the required solution.
View full question & answer→Question 663 Marks
Solve the following equation:
$(\text{e}^\text{y}+1)\cos\text{x dx}+\text{e}^\text{y}\sin\text{x}\text{dy}=0$
Answer$(\text{e}^\text{y}+1)\cos\text{x dx}+\text{e}^\text{y}\sin\text{x}\text{dy}=0$
$(\text{e}^\text{y}+1)\cos\text{x dx}=-\text{e}^\text{y}\sin\text{x}\text{dy}$
$\int\frac{\cos\text{x}}{\sin\text{x}}\ \text{dx}=-\int\frac{\text{e}^\text{y}}{\text{e}^\text{y}+1}\ \text{dy}$
$\int\cot\text{x dx}=-\int\frac{\text{e}^\text{y}}{\text{e}^\text{y}+1}\ \text{dy}$
$\log|\sin\text{x}|=-\log|\text{e}^\text{y}+1|+\log\text|C|$
$\sin\text{x}=\frac{\text{C}}{\text{e}^\text{y}+1}$
$\sin\text{x}(\text{e}^\text{y}+1)=\text{C}$
View full question & answer→Question 673 Marks
Show that $\text{y}=\text{ax}^3+\text{bx}^2+\text{c}$ is a solution of the differential equation $\frac{\text{d}^3\text{y}}{\text{dx}^3}=6\text{a}$
AnswerWe have,
$\text{y}=\text{ax}^3+\text{bx}^2+\text{c}\ ...(1)$
Differentiating both sides of (1) with respect in x, we get
$\frac{\text{dy}}{\text{dx}}=3\text{ax}^2+2\text{bx}\ ...(2)$
Differentiating both sides of (2) with respect in x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=6\text{ax}+2\text{b}\ ...(3)$
Differentiating both sides of (3) with respect in x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=6\text{a}$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 683 Marks
Represent the following families of curves by forming the corresponding differential equation:
$\text{y}=4\text{ax}$
AnswerThe equation of the family of curves is
$\text{y}=4\text{ax}\ ...(1)$
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
$2\text{y}\frac{\text{dt}}{\text{dx}}=4\text{a}$
$\Rightarrow2\text{y}\frac{\text{dy}}{\text{dx}}=\frac{\text{y}^2}{\text{x}}$
$\Rightarrow\text{y}-2\text{x}\frac{\text{dy}}{\text{dx}}=0$
It is the required differential equation.
View full question & answer→Question 693 Marks
show that $\text{y}=\text{be}^\text{x}+\text{ce}^{2\text{x}}$ is a solution of the differential equation, $\frac{\text{d}^2\text{y}}{\text{dx}^2}-3\frac{\text{dy}}{\text{dx}}+2\text{y}=0$
Answer$\text{y}=\text{be}^\text{x}+\text{ce}^{2\text{x}}\ ...(1)$
Differentiating both sides with respect to x,
$\frac{\text{dy}}{\text{dx}}=\text{be}^\text{x}+2\text{ce}^{2\text{x}}\ ...(2)$
Differentiating both sides with respect to x,
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{be}^\text{x}+4\text{ce}^{2\text{x}}\ ...(3)$
now,
$\frac{\text{d}^2\text{y}}{\text{dx}^2}-3\frac{\text{dy}}{\text{dx}}+2\text{y}$
$=\text{be}^\text{x}+4\text{ce}^{2\text{x}}-3(\text{be}^\text{x}+2\text{ce}^{2\text{x}})+2(\text{be}^\text{x}+\text{ce}^{2\text{x}})$
$=\text{be}^\text{x}+4\text{ce}^{2\text{x}}-3\text{be}^\text{x}+6\text{ce}^{2\text{x}}+2\text{be}^\text{x}+2\text{ce}^{2\text{x}}$
$=3\text{be}^\text{x}-3\text{be}^\text{x}+6\text{ce}^{2\text{x}}-6\text{ce}^{2\text{x}}$
$=0$
View full question & answer→Question 703 Marks
Show that $\text{Ax}^2+\text{By}^2=1$ is a solution of the differential equation $\text{x}\Big\{\text{y}=\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dx}}{\text{dy}}\Big)^2\Big\}=\text{y}\frac{\text{dy}}{\text{dx}}.$
AnswerWe have,
$\text{Ax}^2+\text{By}^2=1\ ...(1)$
Differentiating it with respect in x
$2\text{Ax}+2\text{By}\frac{\text{dy}}{\text{dx}}=0$
$\text{y}\frac{\text{dy}}{\text{dx}}=\frac{-2\text{Ax}}{2\text{B}}$
$\text{y}\frac{\text{dy}}{\text{dx}}=\frac{\text{Ax}}{\text{B}}$
Differentiating it with respect in x
$\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=-\frac{\text{A}}{\text{B}}$
$\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\frac{\text{y}}{\text{x}}\frac{\text{dy}}{\text{dx}}$
Using equation (1)
$\text{x}\Big\{\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big\}=\text{y}\frac{\text{dy}}{\text{dx}}$
View full question & answer→Question 713 Marks
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\text{x}^5+\text{x}^2-\frac{2}{\text{x}},\text{x}\ne0$
AnswerWe have, $\frac{\text{dy}}{\text{dx}}=\text{x}^5+\text{x}^2-\frac{2}{\text{x}},\text{x}\ne0$
$\Rightarrow\text{dy}=\Big(\text{x}^5+\text{x}^2-\frac{2}{\text{x}}\Big)\text{dx}$
Intergrating both sides, we get
$\Rightarrow\int\text{dy}=\int\Big(\text{x}^5+\text{x}^2-\frac{2}{\text{x}}\Big)\text{dx}$
$\Rightarrow\text{y}=\frac{\text{x}^6}{6}+\frac{\text{x}^3}{3}-2\log|\text{x}|+\text{C}$
Clearly, $\Rightarrow\text{y}=\frac{\text{x}^6}{6}+\frac{\text{x}^3}{3}-2\log|\text{x}|+\text{C}$ is defined for all $\text{x}\in\text{R}$ except x = 0
Hence, $\Rightarrow\text{y}=\frac{\text{x}^6}{6}+\frac{\text{x}^3}{3}-2\log|\text{x}|+\text{C}$, where $\text{x}\in\text{R}-\{0\},$ is the solution o the given differential equation.
View full question & answer→Question 723 Marks
Form the differential equation having $\text{y}=(\sin^{-1}\text{x})+\text{A}\cos^{-1}\text{x}+\text{B}$ where A and B are aribitrary constants, as its general solution.
Answer$\text{y}=(\sin^{-1}\text{x})+\text{A}\cos^{-1}\text{x}+\text{B}$
$\frac{\text{dy}}{\text{dx}}=2\sin^{-1}\text{x}\times\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)+\text{Ax}\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)=0$
$\sqrt{1-\text{x}^2}\frac{\text{dy}}{\text{dx}}=2\sin^{-1}\text{x}-\text{A}$
$\sqrt{1-\text{x}^2}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\frac{\text{dy}}{\text{dx}}\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)(-2\text{x})\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)-0$
$(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\text{x}\frac{\text{dy}}{\text{dx}}-2=0$
View full question & answer→Question 733 Marks
For each of the differential equations given in find the general solution: $\text{y dx}+(\text{x}-\text{y}^{2})\ \text{dy}=0$
Answer$\text{y dx}+(\text{x}-\text{y}^{2})\text{dy}=0$ $\Rightarrow\frac{\text{dx}}{\text{dy}}=\frac{\text{y}^2-\text{x}}{\text{y}}=\text{y}-\frac{\text{x}}{\text{y}}$
$\Rightarrow\frac{\text{dx}}{\text{dy}}+\frac{\text{x}}{\text{y}}=\text{y}$
This is a linear differential equation of the form:
$\frac{\text{dy}}{\text{dx}}+\text{px}=\text{Q}\ (\text{where p}=\frac{1}{\text{y}\ }\ \text{and}\ \text{Q}=\text{y})$
$\text{Now, I.F}=\text{e}^{\int\text{pdy}}=\text{e}^{\int\frac{1}{\text{y}}\text{dy}}=\text{e}^{\log\text{y}}=\text{y}.$
The general solution of the given differential equation is given by the relation,
$\text{y(I.F)}=\int(\text{Q}\times\text{I.F.})\text{dx}+\text{C}$
$\Rightarrow\text{xy}=\int(\text{y}\cdot\text{y})\text{dy}+\text{C}$
$\Rightarrow\text{xy}=\int\text{y}^2\text{dy}+\text{C}$
$\Rightarrow\text{xy}=\frac{\text{y}^3}{3}+\text{C}$
$\Rightarrow\text{x}=\frac{\text{y}^2}{3}+\frac{\text{C}}{\text{y}}$
View full question & answer→Question 743 Marks
Represent the followinf families of curves by forming the corresponding differential equation:
$\text{y}=\text{e}^{\text{ax}}$
AnswerThe equation of the family of curves is
$\text{y}=\text{e}^{\text{ax}}\ ...(1)$
$\Rightarrow\log\text{y}=\text{ax}$
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{a}$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\log\text{y}}{\text{x}}$
$\Rightarrow\text{x}\frac{\text{dx}}{\text{dx}}=\text{y}\log\text{y}$
It is the required differential equation.
View full question & answer→Question 753 Marks
Show that $\text{y}=\text{Ae}^{\text{bx}}$ is a solution of the differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{y}}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2.$
AnswerWe have,
$\text{y}=\text{Ae}^{\text{bx}}\ ...(1)$
Differentiating both sides of equation (1) with respect to 3, we get
$\frac{\text{dy}}{\text{dx}}=\text{ABe}^{\text{Bx}}\ ...(2)$
Differentiating both sides of equation (2) with respect to 3, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{AB}^2\text{e}^{\text{Bx}}$
$=\frac{(\text{ABe}^{\text{Bx}^2})}{(\text{Ae}^{\text{Bx}})}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{y}}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 763 Marks
For each of the differential equations given in find the general solution:
$\frac{\text{dy}}{\text{dx}}+3\text{y}=\text{e}^{-2\text{x}}$
AnswerThe given differential equation is $\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q}$ (where p = 3 and Q = e -2x) $\text{Now, I.F}=\text{e}^{\int\text{pdx}}=\text{e}^{\int3\text{dx}}=\text{e}^{3\text{x}}.$
The solution of the given differential equation is given by the relation,
$\text{y}(\text{I.F})=\int(\text{Q}\times\text{I.F.})\text{dx}+\text{C}$
$\Rightarrow\ \text{ye}^{\text{3}\text{x}}=\int(\text{e}^{-2\text{x}}\times\text{e}^{3\text{x}})+\text{C}$
$\Rightarrow\ \text{ye}^{\text{3}\text{x}}=\int\text{e}^\text{x}\text{dx}+\text{C}$
$\Rightarrow\ \text{ye}^{\text{3}\text{x}}=\text{e}^\text{x}+\text{C}$
$\Rightarrow\ \text{y}=\text{e}^{\text{-2}\text{x}}+\text{C e}^{\text{-3}\text{x}}$
This is the required general solution of the given differential equation.
View full question & answer→Question 773 Marks
For each of the differential equations given in find the general solution:
$\frac{\text{dy}}{\text{dx}}+\sec\text{xy}=\tan\text{x}\Big(0\leq\text{x}<\frac{\pi}{2}\Big)$
AnswerThe given differential equation is: $\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q (where p}=\sec\text{x and Q}=\tan\text{x})$
$\text{Now, I.F}=\text{e}^{\int\text{pdx}}=\text{e}^{\int\sec\text{x dx}}=\text{e}^{\log(\sec\text{x}+\tan\text{x})}=\sec.\text{x}+\tan\text{x}.$
The general solution of the given differential equation is given by the relation,
$\text{y}(\text{I.F})=\int(\text{Q}\times\text{I.F.})\text{dx}+\text{C}$
$\Rightarrow\ \text{y}(\sec\text{x}+\tan\text{x})=\int\tan\text{x}(\sec\text{x}+\tan\text{x})\text{dx}+\text{C}$
$\Rightarrow\ \text{y}(\sec\text{x}+\tan\text{x})=\int\sec\text{x}\tan\text{x}\text{dx}+\int\tan^2\text{x}\text{dx}+\text{C}$
$\Rightarrow\ \text{y}(\sec\text{x}+\tan\text{x})=\sec\text{x}+\int(\sec^2 \text{x}-1)\text{dx}+\text{C}$
$\Rightarrow\ \text{y}(\sec\text{x}+\tan\text{x})=\sec\text{x}+\tan\text{x}-\text{x}+\text{C}$
This is the required general solution of the given differential equation.
View full question & answer→Question 783 Marks
Represent the following families of curves by forming the corresponding differential equation:
$\text{y}^2=4\text{a}(\text{x}-\text{b})$
AnswerThe equation of the family of curves is
$\text{y}^2=4\text{a}(\text{x}-\text{b})\ ...(1)$
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
$2\text{y}\frac{\text{dy}}{\text{dx}}=4\text{a}$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}=2\text{a}$
Differentiating (2) with respect to x, we get
$\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=0$
It is the required differential equation.
View full question & answer→Question 793 Marks
Differential equation $\text{x}\frac{\text{dy}}{\text{dx}}=1,\text{y}(1)=0$ Function
$\text{y}=\log\text{x}$ AnswerHere, y = logx Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}$
$\text{x}\frac{\text{dy}}{\text{dx}}=1$
so, y=logx is a solution of the equation
If
$\text{x}=1,\text{y}=\log1=0$ so,
y(1) = 0
View full question & answer→Question 803 Marks
Solve the following differential equations: $\frac{\text{dy}}{\text{dx}}+\frac{\cos\text{x}\sin\text{y}}{\cos\text{y}}=0$
Answer$\frac{\text{dy}}{\text{dx}}+\frac{\cos\text{x}\sin\text{y}}{\cos\text{y}}=0$
$\frac{\text{dy}}{\text{dx}}=-\cos\text{x}\tan\text{y}$
$\frac{\text{dy}}{\tan\text{y}}=-\cos\text{x dx}$
$\int\cot\text{ y dy}=-\int\cos\text{x dx}$
$\log|\sin\text{y}|=-\sin\text{x + C}$
$\sin\text{x}+\log|\sin\text{y}|=\text{C}$
View full question & answer→Question 813 Marks
Solve the following differential equations: $\frac{\text{dr}}{\text{dt}}=-\text{rt, r}(0)=\text{r}_{0}$
Answer$\frac{\text{dr}}{\text{dt}}=-\text{rt, r}(0)=\text{r}_{0}$
$\int\frac{\text{dr}}{\text{r}}=-\int\text{tdt}$
$\log|\text{r}|=-\frac{\text{t}^2}{2}+\text{C}...(1)$
Put $\text{t = 0, r = r}_{0}$ inequation (1),
$\log|\text{r}_0|=0+\text{C}$
$\log|\text{r}_0|=\text{C}$
Now,
$\log|\text{r}|=-\frac{\text{t}^2}{2}+\log|\text{r}_0|$
$\frac{\text{r}}{\text{r}_0}=\text{e}^{-\frac{\text{t}^2}{2}}$
$\text{r}=\text{r}_0\text{e}^{-\frac{\text{t}^2}{2}}$
View full question & answer→Question 823 Marks
Solve the following differential equations: $\text{ye}^{\frac{\text{x}}{\text{y}}}\text{dx}=(\text{xe}^{\frac{\text{x}}{\text{y}}}+\text{y}^2)\text{dy, y}\neq0$
Answer$\text{ye}^{\frac{\text{x}}{\text{y}}}\text{dx}=(\text{xe}^{\frac{\text{x}}{\text{y}}}+\text{y}^2)\text{dy}$
$\Rightarrow\text{ye}^{\frac{\text{x}}{\text{y}}}\text{dx = xe}^{\frac{\text{x}}{\text{y}}}\text{dy}+\text{y}^2\text{dy}$
$\Rightarrow\text{ye}^{\frac{\text{x}}{\text{y}}}\text{dx}-\text{xe}^{\frac{\text{x}}{\text{y}}}\text{dy = y}^2\text{dy}$
$\Rightarrow(\text{ydx}-\text{xdy})\text{e}^{\frac{\text{x}}{\text{y}}}=\text{y}^2\text{dy}$
$\Rightarrow\frac{(\text{ydx}-\text{xdy})}{\text{y}^2}\text{e}^{\frac{\text{x}}{\text{y}}}=\text{dy}$
$\Rightarrow\text{e}^{\frac{\text{x}}{\text{y}}}\text{d}\Big(\frac{\text{x}}{\text{y}}\Big)=\text{dy}$
$\Rightarrow\int\text{e}^{\frac{\text{x}}{\text{y}}}\text{d}\Big(\frac{\text{x}}{\text{y}}\Big)=\int\text{dy}$
$\Rightarrow\text{e}^{\frac{\text{x}}{\text{y}}}=\text{y + C}$
View full question & answer→Question 833 Marks
Solve the following differential equations: $\frac{\text{dy}}{\text{dx}}=\text{y}\sin2\text{x, y}(0)=1$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=\text{y}\sin2\text{x, y}(0)=1$
$\Rightarrow\frac{1}{\text{y}}\text{dy}=\sin2\text{x dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}}\text{dy}=\int\sin2\text{x dx}$
$\Rightarrow\log|\text{y}|=-\frac{\cos 2\text{x}}{2}+\text{C}...(1)$
Given: $\text{x}=0,\text{y}=1.$
Substituting the values of x and y in (1), we get
$\log|1|=-\frac{1}{2}+\text{C}$
$\Rightarrow\text{C}=\frac{1}{2}$
Substituting the values of C in (1), we get
$\log|\text{y}|=-\frac{\cos 2\text{x}}{2}+\frac{1}{2}$
$\Rightarrow\log|\text{y}|=\frac{1-\cos 2\text{x}}{2}$
$\Rightarrow\log|\text{y}|=\sin^2\text{x}$
$\Rightarrow\text{y}=\text{e}^{\sin^2\text{x}}$
Hence, $\text{y}=\text{e}^{\sin^2\text{x}}$ is the required solution
View full question & answer→Question 843 Marks
Differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{y}=0,\text{y}(0)=0,\text{y}'(0)=1$ Function
$\text{y}=\sin\text{x}$ AnswerWe have, $\text{y}=\sin{\text{x}} ...(1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\cos\text{x} ...(2)$
Differentiating both sides of (2) with respect to x, we get
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=-\sin{\text{x}}$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=-\text{y}$ [Using (1)]
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{y}=0$
it is the given differential equation.
Here,
$\text{y}=\sin\text{x}$ satisfies the given differential equation; hence, it is a solution. Also, when
$\text{x}=0,\text{y}=\sin0=0,\text{i.e.},\text{y}(0)=0.$ And, when
$\text{x}=0,\text{y}'=\cos 0=1,\text{i.e.,}\text{y}'(0)=1.$ Hence,
$\text{y}=\sin\text{x}$ is the solution to the given initial value problem. View full question & answer→Question 853 Marks
Solve the following differential equations: $(1-\text{x}^2)\text{dy + xy dx = xy}^2\text{ dx}$
Answer$(1-\text{x}^2)\text{dy + xy dx = xy}^2\text{dx}$
$(1-\text{x}^2)\text{dy = dx}(\text{xy}^2-\text{xy})$
$(1-\text{x}^2)\text{dy = xy(y}-1)\text{dx}$
$\int\frac{\text{dy}}{\text{y(y}-1)}=\int\frac{\text{xdx}}{1-\text{x}^2}$
$\int\Big(\frac{1}{\text{y}-1}-\frac{1}{\text{y}}\Big)\text{dy}=\frac{1}{2}\int\frac{2\text{x}}{1-\text{x}^2}\text{dx}$
$\int\Big(\frac{1}{\text{y}-1}-\frac{1}{\text{y}}\Big)\text{dy}=-\frac{1}{2}\int\frac{-2\text{x}}{1-\text{x}^2}\text{dx}$
$\log|\text{y}-1|-\log|\text{y}|=-\frac{1}{2}\log|1-\text{x}^2|+\text{C}$
View full question & answer→Question 863 Marks
Differential equation $\frac{\text{dy}}{\text{dx}}+\text{y}=0,\text{y}(0)=3$ Function
$\text{y}=\text{e}^\text{-x}+2$ AnswerHere, $\text{y}=\text{e}^{\text{x}}+1 ....(1)$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\text{y}-1 ...(2)$
Again, differentiating it with respect to x,
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\frac{\text{dy}}{\text{dx}}$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\frac{\text{dy}}{\text{dx}}=0$
It is given differential equation. so,
y = ex + 1 is a solution of the equation
put x - 0 in equation (1),
⇒ y = e0 + 1 = 2
y(0) = 2
put x = 0 in equation (2),
y' = e0 = 1
y(0) = 1
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