What is the solution for y^{''}-3y^'+2y=x,y(0)=e^x ?

The solution for y^{''}-3y^'+2y=x,y(0)=e^x is y=c_{1}e^{2x}+c_{2}e^x+x/2+3/4

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y^{''}-3y^'+2y=x,y(0)=e^x

y^{^{''}} - 3 y^{^{'}} + 2 y = x, y (0) = e^{x}

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{x}{2} + \frac{3}{4}

Solution steps

y^{''} (x) - 3 y^{'} (x) + 2 y = x

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{x}{2} + \frac{3}{4}

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{x}{2} + \frac{3}{4}

y^{^{''}} + y = 2 cos (x) + x

- 3 y^{^{''}} - 2 y^{^{'}} + y = t^{2} + t - 2 e^{2 t}

x^{^{'''}} - 4 x^{^{''}} + 5 x^{^{'}} - 2 x = 2 t + 3

y^{^{''}} + 4 y = sin (2 x + 3)

y^{^{''}} - 2 y^{^{'}} + y = tan (x)

What is the solution for y^{''}-3y^'+2y=x,y(0)=e^x ?
The solution for y^{''}-3y^'+2y=x,y(0)=e^x is y=c_{1}e^{2x}+c_{2}e^x+x/2+3/4