Find y(e) for x² dy/dx+4xy=4{ x}{x³}x²dy/dx+4xy=4{ x}{x³}, and y(1)=1

Correct Answer is : 3e43e4{3}{e^{4}} {3}{e^{4}}e43e43

Correct Answer is : 3e43e4{3}{e^{4}} {3}{e^{4}}e43e43

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UPSC NDA Math Model Paper 5

Question : 1 of 120

Marks: +1, -0

Find

y(e)

x^{2} \frac{dy}{dx}+4xy=4\frac{\ln x}{x^{3}}x^{2}\frac{dy}{dx}+4xy=4\frac{\ln x}{x^{3}},

y(1)=1

Solution:

Linear differential equation is of first order

x^{2} \frac{dy}{dx}+4xy=4\frac{\ln x}{x^{3}}x^{2}\frac{dy}{dx}+4xy

=4\frac{\ln x}{x^{3}}

\Rightarrow \frac{dy}{dx}+4\frac{y}{x}=4\frac{\ln x}{x^{5}}\frac{dy}{dx}+4\frac{y}{x}=4\frac{\ln x}{x^{5}}

IF=e^{\int \frac{4}{x} \frac{4}{x} dx}

\Rightarrow IF=e^{4\ln x}

\Rightarrow IF=x^{4}

Now,

y \times (IF)=\int Q(

) dx

\Rightarrow y \times x^{4} = \int 4 \frac{\ln x}{x^{5}} \frac{\ln x}{x^{5}} \times x^{4} dx

\Rightarrow y x^{4} = \int 4 \frac{\ln x}{x} \frac{\ln x}{x} dx

Integrating,

\Rightarrow y x^{4} = 2(\ln x)^{2}+c

(where

c

is integration constant)

Given

y(1)=1

\Rightarrow (1)(1)^{4}=2(\ln 1)^{2}+c

\Rightarrow c=1

\therefore y x^{4}=2(\ln x)^{2}+1

For y(e)

y(e)^{4}=2(\ln e)^{2}+1

\Rightarrow y(e^{4})=3

\Rightarrow y=\frac{3}{e^{4}} \frac{3}{e^{4}}

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