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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to 0 - lim (\frac{x + 3}{x})

\frac{d}{d x} (\frac{x ^{4} + 2 x ^{2}}{x ^{2}})

3 t^{2} - 12

y = x^{2}, y = 4 x - x^{2}

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