What is the solution for y^{''}-4y=-7e^{2x}+x,y(0)=1,y^'(0)=3 ?

The solution for y^{''}-4y=-7e^{2x}+x,y(0)=1,y^'(0)=3 is y=(1+3/4)e^{2x}-3/4 e^{-2x}+(-7e^{2x}x-x)/4

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y^{''}-4y=-7e^{2x}+x,y(0)=1,y^'(0)=3

y^{^{''}} - 4 y = - 7 e^{2 x} + x, y (0) = 1, y^{^{'}} (0) = 3

y = (1 + \frac{3}{4}) e^{2 x} - \frac{3}{4} e^{- 2 x} + \frac{- 7 e ^{2 x} x - x}{4}

Solution steps

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

Solve linear ODE:

y = (1 + \frac{3}{4}) e^{2 x} - \frac{3}{4} e^{- 2 x} + \frac{- 7 e ^{2 x} x - x}{4}

y = (1 + \frac{3}{4}) e^{2 x} - \frac{3}{4} e^{- 2 x} + \frac{- 7 e ^{2 x} x - x}{4}

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

y^{^{''}} + 9 y = e^{2 x}

y^{^{''}} + y = 3 + e^{2 x}

y^{^{''}} - y = sin (t), y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 6 y^{^{'}} + 9 y = 2 sin (x)

What is the solution for y^{''}-4y=-7e^{2x}+x,y(0)=1,y^'(0)=3 ?
The solution for y^{''}-4y=-7e^{2x}+x,y(0)=1,y^'(0)=3 is y=(1+3/4)e^{2x}-3/4 e^{-2x}+(-7e^{2x}x-x)/4