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

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

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

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

y = c_{1} e^{3 x} + c_{2} e^{x}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 3 y = 0

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{x}

y = c_{1} e^{3 x} + c_{2} e^{x}

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (- 1) = 6, y^{^{'}} (- 1) = 5

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (- 1) = 6, y^{^{'}} (- 1) = 4

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

\frac{d ^{4} y}{d x ^{4}} - a \frac{d ^{2} y}{d x ^{2}} - b y = 0

x^{^{''}} + 2 x^{^{'}} + 0.75 x = 0

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