What is the solution for xy^'=y+x^2sin(x),y(pi/6)=0 ?

The solution for xy^'=y+x^2sin(x),y(pi/6)=0 is y=-xcos(x)+(sqrt(3)x)/2

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xy^'=y+x^2sin(x),y(pi/6)=0

x y^{^{'}} = y + x^{2} sin (x), y (\frac{π}{6}) = 0

y = - x cos (x) + \frac{3 x}{2}

Solution steps

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

Solve linear ODE:

y = - x cos (x) + \frac{3 x}{2}

y = - x cos (x) + \frac{3 x}{2}

\frac{d}{d x} ((\frac{x ^{2} + 2}{x ^{3} - 1})^{3})

\frac{1}{s ^{3} ( s ^{2} + 1 )}

\frac{d}{d x} (3 x^{2} ln (x) + x^{2})

\int - \frac{1}{sin ( y )} d y

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What is the solution for xy^'=y+x^2sin(x),y(pi/6)=0 ?
The solution for xy^'=y+x^2sin(x),y(pi/6)=0 is y=-xcos(x)+(sqrt(3)x)/2