What is the solution for 2y^{''}-y^'-6y=-7e^{2x},y(0)=5,y^'(0)=2 ?

The solution for 2y^{''}-y^'-6y=-7e^{2x},y(0)=5,y^'(0)=2 is y=3e^{2x}+2e^{-(3x)/2}-e^{2x}x

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2y^{''}-y^'-6y=-7e^{2x},y(0)=5,y^'(0)=2

2 y^{^{''}} - y^{^{'}} - 6 y = - 7 e^{2 x}, y (0) = 5, y^{^{'}} (0) = 2

y = 3 e^{2 x} + 2 e^{- \frac{3 x}{2}} - e^{2 x} x

Solution steps

2 y^{''} (x) - y^{'} (x) - 6 y = - 7 e^{2 x}

Solve linear ODE:

y = 3 e^{2 x} + 2 e^{- \frac{3 x}{2}} - e^{2 x} x

y = 3 e^{2 x} + 2 e^{- \frac{3 x}{2}} - e^{2 x} x

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

y^{^{''}} + 3 y^{^{'}} - 10 = 0, y (0) = 0, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d x ^{2}} + \frac{d}{d x} (8 y) + 65 y = 100

y^{^{''}} + 2 y^{^{'}} - 15 y = - 5 x^{2} + 3 x - 3

a y^{^{''}} + b y = c

What is the solution for 2y^{''}-y^'-6y=-7e^{2x},y(0)=5,y^'(0)=2 ?
The solution for 2y^{''}-y^'-6y=-7e^{2x},y(0)=5,y^'(0)=2 is y=3e^{2x}+2e^{-(3x)/2}-e^{2x}x