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Differential Equations

Section: Mathematics

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Question : 1 of 48

Marks: +1, -0

If

x \log x \;\frac{dy}{dx}+y=\log x^2

y(e)=0

y(e^2)=

[AP EAPCET 24 May 2025 S1]

0
1
$\frac{1}{2}$
$\frac{3}{2}$

Solution:

Given,

x \log x \;\frac{dy}{dx}+y=\log x^2

y(e)=0

\Rightarrow \frac{dy}{dx}+\frac{1}{x \log x}y=\frac{\log x^2}{x \log x}

\Rightarrow \frac{dy}{dx}+\frac{1}{x \log x}y=\frac{2 \log x}{x \log x}=\frac{2}{x}

This is a linear differential equation of the form

\frac{dy}{dx}+P(x)y=Q(x)

,

Where

P(x)=\frac{1}{x \log x}

and

Q(x)=\frac{2}{x}

Now,

IF = e^{\int \frac{1}{x \log x} dx}

Let

u=\log x

.

Then

du = \frac{1}{x} dx

So,

IF = e^{\int \frac{1}{u} du} = e^{\log u} = u = \log x

So, the solution is given by

y \cdot IF = \int Q(x) \cdot IF dx + c

y \log x = \int \frac{2}{x} \log x dx + c

Let

v=\log x

Then,

dv = \frac{1}{x} dx

\text{So,}\; y \log x = \int 2v dv + c = v^2 + c

\Rightarrow y \log x = v^2 + c = (\log x)^2 + c

Given,

y(e)=0

So,

y \log x = (\log x)^2 + c

\Rightarrow 0 \cdot \log(e) = (\log e)^2 + c

\Rightarrow 0 = 1 + c \Rightarrow c = -1

∴ The particular solution is

y \log x = (\log x)^2 - 1

\Rightarrow y = \frac{(\log x)^2 - 1}{\log x}

So,

y(e^2) = \frac{(\log e^2)^2 - 1}{\log(e^2)}

= \frac{(2 \log e)^2 - 1}{2 \log(e)} \quad [\because \log e = 1]

= \frac{2^2 - 1}{2} = \frac{4 - 1}{2} = \frac{3}{2}

\therefore y(e^2) = \frac{3}{2}

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