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

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

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y^{''}-3y^'-4y=2^{t-2},y(2)=1,y^'(2)=-1

y^{^{''}} - 3 y^{^{'}} - 4 y = 2^{t - 2}, y (2) = 1, y^{^{'}} (2) = - 1

y = \frac{e ^{4 t - 10} ( 5 e ^{2} ln ( 2 e ) + e ^{2} - 5 ln ( 2 e ) c _{2} - 5 e ^{10} ln ( 2 e ) \cdot \int \frac{1}{5 e ^{8}} d 2 )}{5 ln ( 2 e )} + c_{2} e^{- t} + e^{4 t} \cdot \int \frac{2 ^{t - 2}}{5 e ^{4 t}} d t - \frac{2 ^{t - 2}}{5 ln ( 2 e )}

Solution steps

y^{''} (t) - 3 y^{'} (t) - 4 y = 2^{t - 2}

Solve linear ODE:

y = \frac{e ^{4 t - 10} ( 5 e ^{2} ln ( 2 e ) + e ^{2} - 5 ln ( 2 e ) c _{2} - 5 e ^{10} ln ( 2 e ) \cdot \int \frac{1}{5 e ^{8}} d 2 )}{5 ln ( 2 e )} + c_{2} e^{- t} + e^{4 t} \cdot \int \frac{2 ^{t - 2}}{5 e ^{4 t}} d t - \frac{2 ^{t - 2}}{5 ln ( 2 e )}

y = \frac{e ^{4 t - 10} ( 5 e ^{2} ln ( 2 e ) + e ^{2} - 5 ln ( 2 e ) c _{2} - 5 e ^{10} ln ( 2 e ) \cdot \int \frac{1}{5 e ^{8}} d 2 )}{5 ln ( 2 e )} + c_{2} e^{- t} + e^{4 t} \cdot \int \frac{2 ^{t - 2}}{5 e ^{4 t}} d t - \frac{2 ^{t - 2}}{5 ln ( 2 e )}

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

(d^{2} - 2 d + 2) y = (x + e^{x}) sin (x)

y^{^{''}} - y = \frac{2 e ^{2 x}}{e ^{2 x} + 1}

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + y = 4 sin (x)

y^{^{''}} - 4 y^{^{'}} + 3 y = e^{x}

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