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

The solution for y^{''}-2y^'+y=t+1 is y=c_{1}e^t+c_{2}te^t+t+3

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

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

y = c_{1} e^{t} + c_{2} t e^{t} + t + 3

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} t e^{t} + t + 3

y = c_{1} e^{t} + c_{2} t e^{t} + t + 3

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

\frac{d}{d x} (- 5 cos (cos (x^{6})))

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

\frac{d}{d x} (\frac{ln ( 1 + x )}{x})

x \to 0 lim (h (x) (x^{\frac{2}{3}}))

What is the solution for y^{''}-2y^'+y=t+1 ?
The solution for y^{''}-2y^'+y=t+1 is y=c_{1}e^t+c_{2}te^t+t+3