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

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

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

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

y = - \frac{t ^{3} cos ( 4 t )}{4} - \frac{t ^{3}}{4}

Solution steps

t y^{'} (t) = 3 y + t^{4} sin (4 t)

Solve linear ODE:

y = - \frac{t ^{3} cos ( 4 t )}{4} - \frac{t ^{3}}{4}

y = - \frac{t ^{3} cos ( 4 t )}{4} - \frac{t ^{3}}{4}

\frac{d}{d x} (sec (π x))

\int \frac{4 x + 2}{x + 2} d x

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

x y d x + (x^{2} + y^{2}) d y = 0

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