What is the solution for y^{''}+4y=t^2+10e^t,y(0)=0,y^'(0)=7 ?

The solution for y^{''}+4y=t^2+10e^t,y(0)=0,y^'(0)=7 is y=-15/8 cos(2t)+5/2 sin(2t)+(t^2)/4-1/8+2e^t

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y^{''}+4y=t^2+10e^t,y(0)=0,y^'(0)=7

y^{^{''}} + 4 y = t^{2} + 10 e^{t}, y (0) = 0, y^{^{'}} (0) = 7

y = - \frac{15}{8} cos (2 t) + \frac{5}{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + 2 e^{t}

Solution steps

y^{''} (t) + 4 y = t^{2} + 10 e^{t}

Solve linear ODE:

y = - \frac{15}{8} cos (2 t) + \frac{5}{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + 2 e^{t}

y = - \frac{15}{8} cos (2 t) + \frac{5}{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + 2 e^{t}

What is the solution for y^{''}+4y=t^2+10e^t,y(0)=0,y^'(0)=7 ?
The solution for y^{''}+4y=t^2+10e^t,y(0)=0,y^'(0)=7 is y=-15/8 cos(2t)+5/2 sin(2t)+(t^2)/4-1/8+2e^t