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

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

y^{^{''}} - 3 y^{^{'}} + 2 y = x, y^{1} (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^{^{''}} - \frac{4}{9} y = x^{3} - 2

x^{^{''}} - 4 x^{^{'}} + 4 x = 84 t^{2} e^{2 t}

y^{^{''}} + 25 y = 25, y (0) = 1, y^{^{'}} (0) = 5

y^{^{''}} - 2 y^{^{'}} - 3 y = e^{4 x}

y^{^{''}} + 10 y^{^{'}} + 25 y = - 8 - 5 x + 25 x^{2}

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