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

The solution for y^{''}-4y^'=3e^{2x},y(0)=0,y^'(0)=0 is y=(-3/8+3/4)e^{4x}+3/8-3/4 e^{2x}

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

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

y = (- \frac{3}{8} + \frac{3}{4}) e^{4 x} + \frac{3}{8} - \frac{3}{4} e^{2 x}

Solution steps

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

Solve linear ODE:

y = (- \frac{3}{8} + \frac{3}{4}) e^{4 x} + \frac{3}{8} - \frac{3}{4} e^{2 x}

y = (- \frac{3}{8} + \frac{3}{4}) e^{4 x} + \frac{3}{8} - \frac{3}{4} e^{2 x}

What is the solution for y^{''}-4y^'=3e^{2x},y(0)=0,y^'(0)=0 ?
The solution for y^{''}-4y^'=3e^{2x},y(0)=0,y^'(0)=0 is y=(-3/8+3/4)e^{4x}+3/8-3/4 e^{2x}