What is the solution for xy^'=y+x^2sin(x),y(pi/4)=0 ?

The solution for xy^'=y+x^2sin(x),y(pi/4)=0 is y=-xcos(x)+x/(sqrt(2))

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xy^'=y+x^2sin(x),y(pi/4)=0

x y^{^{'}} = y + x^{2} sin (x), y (\frac{π}{4}) = 0

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{1}{2} e^{\frac{1}{2} x})

x \to + 3 lim (\frac{8}{x - 3})

5 x, \frac{5}{x}, 0, 5

\int_{0}^{\frac{π}{4}} 6 tan^{3} (x) d x

\frac{\partial}{\partial y} ((2 x - y + 5)^{2})

What is the solution for xy^'=y+x^2sin(x),y(pi/4)=0 ?
The solution for xy^'=y+x^2sin(x),y(pi/4)=0 is y=-xcos(x)+x/(sqrt(2))