SATHEE: Differential Equations 2 Question 6

Differential Equations 2 Question 6

6. The solution of the differential equation $x \frac{d y}{d x} + 2 y = x^{2} (x \neq 0)$ with $y (1) = 1$ , is (2019 Main, 9 April I)

(a) $y = \frac{x^{2}}{4} + \frac{3}{4 x^{2}}$

(b) $y = \frac{x^{3}}{5} + \frac{1}{5 x^{2}}$

(d) $y = \frac{4}{5} x^{3} + \frac{1}{5 x^{2}}$

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Answer:

Correct Answer: 6. (a)

Solution:

Given differential equation is

$\begin{aligned} x \frac{d y}{d x} + 2 y = x^{2}, (x \neq 0) \\ \Rightarrow \frac{d y}{d x} + \frac{2}{x} y & = x, \end{aligned}$

which is a linear differential equation of the form

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

Here, $P = \frac{2}{x}$ and $Q = x$

$∴ I F = e^{\int \frac{2}{x} d x} = e^{2 \log x} = x^{2}$

Since, solution of the given differential equation is

$\begin{aligned} y \times I F = \int (Q \times I F) d x + C \\ ∴ & y (x^{2}) = \int (x \times x^{2}) d x + C \Rightarrow y x^{2} = \frac{x^{4}}{4} + C \\ ∵ & y (1) = 1, so 1 = \frac{1}{4} + C \Rightarrow C = \frac{3}{4} \\ ∴ & y x^{2} = \frac{x^{4}}{4} + \frac{3}{4} \Rightarrow y = \frac{x^{2}}{4} + \frac{3}{4 x^{2}} \end{aligned}$

Differential Equations 2 Question 6

6. The solution of the differential equation $x \frac{d y}{d x} + 2 y = x^{2} (x \neq 0)$ with $y (1) = 1$ , is (2019 Main, 9 April I)

Answer:

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Differential Equations 2 Question 6

6. The solution of the differential equation xdydx+2y=x2(x≠0) with y(1)=1, is (2019 Main, 9 April I)

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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6. The solution of the differential equation $x \frac{d y}{d x} + 2 y = x^{2} (x \neq 0)$ with $y (1) = 1$ , is (2019 Main, 9 April I)