What is the solution for xy^'=y+x^2sin(x),y(pi/2)=0 ?

The solution for xy^'=y+x^2sin(x),y(pi/2)=0 is y=-xcos(x)

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xy^'=y+x^2sin(x),y(pi/2)=0

x y^{^{'}} = y + x^{2} sin (x), y (\frac{π}{2}) = 0

y = - x cos (x)

Solution steps

x y^{'} (x) = y + x^{2} sin (x)

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y = - x cos (x)

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x \cdot \frac{d y}{d x} - y = x^{2} + y^{2}

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\frac{d}{d t} (\frac{2 t}{2 t ^{2} + 1})

What is the solution for xy^'=y+x^2sin(x),y(pi/2)=0 ?
The solution for xy^'=y+x^2sin(x),y(pi/2)=0 is y=-xcos(x)