What is the solution for y^{''}-2y^'+y=2cos(t)+4sin(t),y(0)=3,y^'(0)=2 ?

The solution for y^{''}-2y^'+y=2cos(t)+4sin(t),y(0)=3,y^'(0)=2 is y=e^t+2e^tt-sin(t)+2cos(t)

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y^{''}-2y^'+y=2cos(t)+4sin(t),y(0)=3,y^'(0)=2

y^{^{''}} - 2 y^{^{'}} + y = 2 cos (t) + 4 sin (t), y (0) = 3, y^{^{'}} (0) = 2

y = e^{t} + 2 e^{t} t - sin (t) + 2 cos (t)

Solution steps

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

Solve linear ODE:

y = e^{t} + 2 e^{t} t - sin (t) + 2 cos (t)

y = e^{t} + 2 e^{t} t - sin (t) + 2 cos (t)

\frac{d}{d x} (x^{6})

\frac{d}{d x} (\frac{1}{e ^{2 x}})

s l o p e y = 6 x + x^{2}

\int \frac{1}{t ^{2} 6 - t ^{2}} d t

x \to 0 lim (\frac{tan ( 8 x )}{x})

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