For the equation \;dy/dx+7x²y=0, if y(0)=3/7, then the value of y(1) is

Correct Answer is : 37e−73\;3/7e^{-7/3}73e−37

Correct Answer is : 37e−73\;3/7e^{-7/3}73e−37

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GATE Mechanical Engineering (ME) 2019 Shift 1 Solved Paper

Question : 30 of 65

Marks: +1, -0

\;\frac{dy}{dx}+7x^{2}y=0

y(0)=\frac{3}{7}

y(1)

Solution:

The given equation is

\;\frac{dy}{dx}+7x^{2}y=0

Or,

\;\frac{dy}{dx}=-7x^{2}y

\;\frac{dy}{y}=-7x^{2}dx

Integrating both sides,

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

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

Using

y(0)=\frac{3}{7}

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

Or,

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

Therefore,

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

Or,

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

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

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

Therefore,

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

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