What is the solution for y^'=-5y+5t^2+2t,y(0)= 1/3 ?

The solution for y^'=-5y+5t^2+2t,y(0)= 1/3 is y=t^2+1/(3e^{5t)}

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y^'=-5y+5t^2+2t,y(0)= 1/3

y^{^{'}} = - 5 y + 5 t^{2} + 2 t, y (0) = \frac{1}{3}

y = t^{2} + \frac{1}{3 e ^{5 t}}

Solution steps

y^{'} (t) = - 5 y + 5 t^{2} + 2 t

Solve linear ODE:

y = t^{2} + \frac{1}{3 e ^{5 t}}

y = t^{2} + \frac{1}{3 e ^{5 t}}

\int \frac{3}{4 - x ^{2}} d x

\frac{d}{d θ} (θ - sin (θ))

x \to 0 lim (\frac{1}{sin ( x )})

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

\frac{d}{d x} (6 x^{2} - x^{3})

What is the solution for y^'=-5y+5t^2+2t,y(0)= 1/3 ?
The solution for y^'=-5y+5t^2+2t,y(0)= 1/3 is y=t^2+1/(3e^{5t)}