For the equation \frac{d y}{d x}+7 x^{2} y=0, if y(0)=3 \/ 7, then the value of y(1) is

Correct Answer is : 37e−7∕3

GATE Mechanical Engineering (ME) 2019 Shift 1 Solved Paper

Question : 30 of 65

Marks: +1, -0

For the equation

\;{d y}/{d x}+7 x^{2} y=0

, if

y(0)=3 \/ 7

, then the value of

y(1)

Solution:

The given equation is

\;{d y}/{d x}+7 x^{2} y=0

,
Or,

\;{d y}/{d x}=-7 x^{2} y

\;{d y}/{y}=-7 x^{2} d x

Integrating both sides,

\ln y=-\;{7}/{3} x^{3}+c

y=e^{(-\;{7}/{3} x^{3}+c)}

Using

y(0)=3{/} 7

3 {/} 7=e^{(0+C)}=e^{c}

Or,

C=\ln \;{3}/{7}

Therefore,

\ln y=-\;{7}/{3} x^{3}+\ln \;{3}/{7}

Or,

\ln y-\ln \;{3}/{7}=-\;{7}/{3} x^{3}

\ln \;{7 y}/{3}=-\;{7}/{3} x^{3}

\;{7 y}/{3}=e^{-\;{2}/{3} x^{2}}

Therefore,

y(1)=\;{3}/{7} e^{-\;{2}/{2}}

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