SATHEE: Differential Equations 2 Question 7

Differential Equations 2 Question 7

7. Let $y = y (x)$ be the solution of the differential equation, ${(x^{2} + 1)}^{2} \frac{d y}{d x} + 2 x (x^{2} + 1) y = 1$ such that $y (0) = 0$ . If $\sqrt{a} y (1) = \frac{π}{32}$ , then the value of ’ $a$ ’ is (2019 Main, 8 April I)

(a) $\frac{1}{4}$

(b) $\frac{1}{2}$

(d) $\frac{1}{16}$

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Answer:

Correct Answer: 7. (d)

Solution:

Given differential equation is

$\begin{aligned} {(x^{2} + 1)}^{2} \frac{d y}{d x} + 2 x (x^{2} + 1) y = 1 \\ \Rightarrow & \frac{d y}{d x} + \frac{2 x}{1 + x^{2}} y = \frac{1}{{(1 + x^{2})}^{2}} \end{aligned}$

$[dividing each term by {(1 + x^{2})}^{2}]$

This is a linear differential equation of the form

$\frac{d y}{d x} + P \cdot y = Q$

Here, $P = \frac{2 x}{(1 + x^{2})}$ and $Q = \frac{1}{{(1 + x^{2})}^{2}}$

$∴$ Integrating Factor (IF) $= e^{\int \frac{2 x}{1 + x^{2}} d x}$

$= e^{\ln (1 + x^{2})} = (1 + x^{2})$

and required solution of differential Eq. (i) is given by

$\begin{aligned} y \cdot (I F) & = \int Q (I F) d x + C \\ \Rightarrow & y (1 + x^{2}) & = \int \frac{1}{{(1 + x^{2})}^{2}} (1 + x^{2}) d x + C \\ \Rightarrow & y (1 + x^{2}) & = \int \frac{d x}{1 + x^{2}} + C \\ \Rightarrow & y (1 + x^{2}) & = \tan^{- 1} (x) + C \\ ∵ & y (0) & = 0 \\ ∴ & C \\ ∴ & y (1 + x^{2}) & = \tan^{- 1} x \\ \Rightarrow & y & = \frac{\tan^{- 1} x}{1 + x^{2}} \\ \Rightarrow \end{aligned}$

Now, at $x = 1$

[multiplying both sides by $\sqrt{a}$ ]

$\begin{aligned} \sqrt{a} y (1) & = \sqrt{a} \frac{\tan^{- 1} (1)}{1 + 1} = \sqrt{a} \frac{\frac{π}{4}}{2} = \frac{\sqrt{a} π}{8} = \frac{π}{32} (given) \\ ∴ \sqrt{a} & = \frac{1}{4} \Rightarrow a = \frac{1}{16} \end{aligned}$

Differential Equations 2 Question 7

7. Let $y = y (x)$ be the solution of the differential equation, ${(x^{2} + 1)}^{2} \frac{d y}{d x} + 2 x (x^{2} + 1) y = 1$ such that $y (0) = 0$ . If $\sqrt{a} y (1) = \frac{π}{32}$ , then the value of ’ $a$ ’ is (2019 Main, 8 April I)

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Differential Equations 2 Question 7

7. Let y=y(x) be the solution of the differential equation, (x2+1)2dydx+2x(x2+1)y=1 such that y(0)=0. If ay(1)=π32, then the value of ’ a ’ is (2019 Main, 8 April I)

Answer:

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Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

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7. Let $y = y (x)$ be the solution of the differential equation, ${(x^{2} + 1)}^{2} \frac{d y}{d x} + 2 x (x^{2} + 1) y = 1$ such that $y (0) = 0$ . If $\sqrt{a} y (1) = \frac{π}{32}$ , then the value of ’ $a$ ’ is (2019 Main, 8 April I)