What is the solution for ty^'=2y+3t^3cos(2t),y(pi/4)=0 ?

The solution for ty^'=2y+3t^3cos(2t),y(pi/4)=0 is y=(3t^2sin(2t))/2-(3t^2)/2

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ty^'=2y+3t^3cos(2t),y(pi/4)=0

t y^{^{'}} = 2 y + 3 t^{3} cos (2 t), y (\frac{π}{4}) = 0

y = \frac{3 t ^{2} sin ( 2 t )}{2} - \frac{3 t ^{2}}{2}

Solution steps

t y^{'} (t) = 2 y + 3 t^{3} cos (2 t)

Solve linear ODE:

y = \frac{3 t ^{2} sin ( 2 t )}{2} - \frac{3 t ^{2}}{2}

y = \frac{3 t ^{2} sin ( 2 t )}{2} - \frac{3 t ^{2}}{2}

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What is the solution for ty^'=2y+3t^3cos(2t),y(pi/4)=0 ?
The solution for ty^'=2y+3t^3cos(2t),y(pi/4)=0 is y=(3t^2sin(2t))/2-(3t^2)/2