What is the solution for y^{''}-4y^'+3y=2e^{5x},y(0)=1,y(0)=-2 ?

The solution for y^{''}-4y^'+3y=2e^{5x},y(0)=1,y(0)=-2 is y=(-c_{2}+3/4)e^{3x}+c_{2}e^x+1/4 e^{5x}

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y^{''}-4y^'+3y=2e^{5x},y(0)=1,y(0)=-2

y^{^{''}} - 4 y^{^{'}} + 3 y = 2 e^{5 x}, y (0) = 1, y (0) = - 2

y = (- c_{2} + \frac{3}{4}) e^{3 x} + c_{2} e^{x} + \frac{1}{4} e^{5 x}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 3 y = 2 e^{5 x}

Solve linear ODE:

y = (- c_{2} + \frac{3}{4}) e^{3 x} + c_{2} e^{x} + \frac{1}{4} e^{5 x}

y = (- c_{2} + \frac{3}{4}) e^{3 x} + c_{2} e^{x} + \frac{1}{4} e^{5 x}

y^{^{''}} + 2 y^{^{'}} + 5 y = e^{x} sin (x)

y^{^{''}} + y = \frac{1}{sin ^{2} ( x )}

5 y^{^{''}} + 10 y^{^{'}} = 3 x^{2}

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

y^{^{''}} + 4 y = 4 sec^{2} (2 x)

What is the solution for y^{''}-4y^'+3y=2e^{5x},y(0)=1,y(0)=-2 ?
The solution for y^{''}-4y^'+3y=2e^{5x},y(0)=1,y(0)=-2 is y=(-c_{2}+3/4)e^{3x}+c_{2}e^x+1/4 e^{5x}