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

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

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y^{''}-2y+y=t+1,y(0)=0,y^'(0)=0

y^{^{''}} - 2 y + y = t + 1, y (0) = 0, y^{^{'}} (0) = 0

y = - t + e^{t} - 1

Solution steps

y^{''} (t) - 2 y + y = t + 1

Solve linear ODE:

y = - t + e^{t} - 1

y = - t + e^{t} - 1

f (x) = - \frac{2}{x}

\frac{\partial}{\partial x} (4 x e^{4 x y})

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

\int_{1}^{2} 2 (1 - \frac{1}{x ^{2}}) d x

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