What is the solution for y^{''}-2y^'+y=te^t+4,y(0)=1,y^'(0)=1 ?

The solution for y^{''}-2y^'+y=te^t+4,y(0)=1,y^'(0)=1 is y=-3e^t+4e^tt+(e^tt^3)/6+4

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y^{''}-2y^'+y=te^t+4,y(0)=1,y^'(0)=1

y^{^{''}} - 2 y^{^{'}} + y = t e^{t} + 4, y (0) = 1, y^{^{'}} (0) = 1

y = - 3 e^{t} + 4 e^{t} t + \frac{e ^{t} t ^{3}}{6} + 4

Solution steps

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

Solve linear ODE:

y = - 3 e^{t} + 4 e^{t} t + \frac{e ^{t} t ^{3}}{6} + 4

y = - 3 e^{t} + 4 e^{t} t + \frac{e ^{t} t ^{3}}{6} + 4

f (x) = x^{4} (3 - x)^{3}

θ cos (θ) + sin (θ)

\int_{6}^{8} \frac{98}{( x - 6 ) ^{3}} d x

\frac{d}{d x} (7 x^{3} - 5 x + 8)

x \to 2 lim (- 1.999 + 2)

What is the solution for y^{''}-2y^'+y=te^t+4,y(0)=1,y^'(0)=1 ?
The solution for y^{''}-2y^'+y=te^t+4,y(0)=1,y^'(0)=1 is y=-3e^t+4e^tt+(e^tt^3)/6+4