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

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

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

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

y = \frac{1}{4} e^{4 t} - \frac{1}{4}

Solution steps

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

Solve linear ODE:

y = \frac{1}{4} e^{4 t} - \frac{1}{4}

y = \frac{1}{4} e^{4 t} - \frac{1}{4}

y^{^{''}} + 36 y = 0, y (\frac{π}{6}) = 1, y^{^{'}} (\frac{π}{6}) = - 2

y^{^{'''}} - y^{^{''}} - 24 y^{^{'}} - 36 y = 0

\frac{d ^{2} y}{d x ^{2}} + 225 y = 0

y^{^{'''}} - 27 y = 0, y (0) = - 1, y^{^{'}} (0) = 6, y^{^{'''}} (0) = 18

y^{^{'''}} - 8 y = 0, y (0) = 1, y^{^{'}} (0) = - 1, y^{^{''}} (0) = 0

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