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

The solution for y^{''}-4y^'+(4-1)y=0,y(0)=0,y^'(0)=1 is y= 1/2 e^{3t}-1/2 e^t

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

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

y = \frac{1}{2} e^{3 t} - \frac{1}{2} e^{t}

Solution steps

y^{''} (t) - 4 y^{'} (t) + (4 - 1) y = 0

Solve linear ODE:

y = \frac{1}{2} e^{3 t} - \frac{1}{2} e^{t}

y = \frac{1}{2} e^{3 t} - \frac{1}{2} e^{t}

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} + 6 y = 0

- y^{^{''}} + 9 y = 0

25 x^{^{''}} + 50 x^{^{'}} + 169 x = 0, x (0) = 2, x^{^{'}} (0) = 6

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

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

What is the solution for y^{''}-4y^'+(4-1)y=0,y(0)=0,y^'(0)=1 ?
The solution for y^{''}-4y^'+(4-1)y=0,y(0)=0,y^'(0)=1 is y= 1/2 e^{3t}-1/2 e^t