What is the solution for y^{''}+4y=e^2t ?

The solution for y^{''}+4y=e^2t is y=c_{1}cos(2t)+c_{2}sin(2t)+(e^2t)/4

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y^{''}+4y=e^2t

y^{^{''}} + 4 y = e^{2} t

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{e ^{2} t}{4}

Solution steps

y^{''} (t) + 4 y = e^{2} t

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{e ^{2} t}{4}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{e ^{2} t}{4}

(d^{2} + 10 d + 25) y = e^{- 5 x}

(d^{2} - 3 d + 2) \cdot y = e^{2 x}

(d^{2} - 4) \cdot y = x \cdot sinh (x)

(d^{2} - 2 d + 3) \cdot y = cos (x) + x^{2}

(d^{2} - 4 d + 4) y = e^{2} x^{2}

What is the solution for y^{''}+4y=e^2t ?
The solution for y^{''}+4y=e^2t is y=c_{1}cos(2t)+c_{2}sin(2t)+(e^2t)/4