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

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

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

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

y = - e^{2 x} + 2 e^{x} - e^{x} x

Solution steps

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

Solve linear ODE:

y = - e^{2 x} + 2 e^{x} - e^{x} x

y = - e^{2 x} + 2 e^{x} - e^{x} x

y^{^{''}} + 4 y = 3 sin (t)

y^{^{''}} + 100 y^{^{'}} + 2500 y = 30

3 y^{^{''}} - 5 y^{^{'}} - 2 y = 3 e^{2 x}

4 y^{^{''}} - 8 y^{^{'}} = cos (2 x) + 7 x^{2}

y^{^{''}} - 2 y^{^{'}} + y = \frac{4 e ^{x}}{x ^{2} + 1}

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