SATHEE: Differential Equations 2 Question 13

Differential Equations 2 Question 13

13. If $y = y (x)$ is the solution of the differential equation, $x \frac{d y}{d x} + 2 y = x^{2}$ satisfying $y (1) = 1$ , then $y \frac{1}{2}$ is equal to

(a) $\frac{13}{16}$

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

(d) $\frac{7}{64}$

(2019 Main, 9 Jan I)

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Answer:

Correct Answer: 13. (c)

Solution:

Given differential equation can be rewritten as $\frac{d y}{d x} + \frac{2}{x} \cdot y = x$ , which is a linear differential equation of the form $\frac{d y}{d x} + P y = Q$ , where $P = \frac{2}{x}$ and $Q = x$ .

Now, integrating factor

$[∵ e^{\log f (x)} = f (x)]$

$(IF) = e^{\int \frac{2}{x} d x} = e^{2 \log x} = e^{\log x^{2}} = x^{2}$

and the solution is given by

$\begin{aligned} y (I F) & = \int (Q \times I F) d x + C \\ \Rightarrow y x^{2} & = \int x^{3} d x + C \\ \Rightarrow y x^{2} & = \frac{x^{4}}{4} + C \end{aligned}$

Since, it is given that $y = 1$ when $x = 1$

$∴$ From Eq. (i), we get

$\begin{aligned} 1 & = \frac{1}{4} + C \Rightarrow C = \frac{3}{4} \\ ∴ & 4 x^{2} y & = x^{4} + 3 [using Eqs. (i) and (ii)] \\ \Rightarrow y & = \frac{x^{4} + 3}{4 x^{2}} \\ Now, & y & \frac{1}{2} & = \frac{\frac{1}{16} + 3}{4 \times \frac{1}{4}} = \frac{49}{16} \end{aligned}$

Differential Equations 2 Question 13

13. If $y = y (x)$ is the solution of the differential equation, $x \frac{d y}{d x} + 2 y = x^{2}$ satisfying $y (1) = 1$ , then $y \frac{1}{2}$ is equal to

Answer:

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Differential Equations 2 Question 13

13. If y=y(x) is the solution of the differential equation, xdydx+2y=x2 satisfying y(1)=1, then y12 is equal to

Answer:

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13. If $y = y (x)$ is the solution of the differential equation, $x \frac{d y}{d x} + 2 y = x^{2}$ satisfying $y (1) = 1$ , then $y \frac{1}{2}$ is equal to