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

The solution for y^{''}-3y^'+2y=1,y^'(4)=3 is y=(e^{2t-8}(3-e^4c_{2}))/2+c_{2}e^t+1/2

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y^{''}-3y^'+2y=1,y^'(4)=3

y^{^{''}} - 3 y^{^{'}} + 2 y = 1, y^{^{'}} (4) = 3

y = \frac{e ^{2 t - 8} ( 3 - e ^{4} c _{2} )}{2} + c_{2} e^{t} + \frac{1}{2}

Solution steps

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

Solve linear ODE:

y = \frac{e ^{2 t - 8} ( 3 - e ^{4} c _{2} )}{2} + c_{2} e^{t} + \frac{1}{2}

y = \frac{e ^{2 t - 8} ( 3 - e ^{4} c _{2} )}{2} + c_{2} e^{t} + \frac{1}{2}

y^{^{''}} + 2 y^{^{'}} - 3 y = e^{- x} (- 4 x + 3)

y^{^{''}} + 2 y^{^{'}} - 8 y = 36 e^{2 t}

y^{^{''}} + 4 y^{^{'}} + 3 y = - e^{- 6 t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 6 y^{^{'}} - 7 y = 11 e^{2 x}, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} - y^{^{'}} - 6 = 0

What is the solution for y^{''}-3y^'+2y=1,y^'(4)=3 ?
The solution for y^{''}-3y^'+2y=1,y^'(4)=3 is y=(e^{2t-8}(3-e^4c_{2}))/2+c_{2}e^t+1/2