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

The solution for y^{''}-4y^'=1,y(0)=1,y^'(0)=4 is y=(1+1/16)e^{4t}-1/16-t/4

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

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

y = (1 + \frac{1}{16}) e^{4 t} - \frac{1}{16} - \frac{t}{4}

Solution steps

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

Solve linear ODE:

y = (1 + \frac{1}{16}) e^{4 t} - \frac{1}{16} - \frac{t}{4}

y = (1 + \frac{1}{16}) e^{4 t} - \frac{1}{16} - \frac{t}{4}

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

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

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

y^{^{''}} + y = - 16 cos (3 x)

4 y^{^{''}} - y = e^{\frac{x}{2}} + 7

What is the solution for y^{''}-4y^'=1,y(0)=1,y^'(0)=4 ?
The solution for y^{''}-4y^'=1,y(0)=1,y^'(0)=4 is y=(1+1/16)e^{4t}-1/16-t/4