What is the solution for y^{''}=32e^{2t},y(0)=3,y^'(0)=8 ?

The solution for y^{''}=32e^{2t},y(0)=3,y^'(0)=8 is y=8e^{2t}-8t-5

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y^{''}=32e^{2t},y(0)=3,y^'(0)=8

y^{^{''}} = 32 e^{2 t}, y (0) = 3, y^{^{'}} (0) = 8

y = 8 e^{2 t} - 8 t - 5

Solution steps

y^{''} (t) = 32 e^{2 t}

Solve

y^{''} (t) = 32 e^{2 t} : y^{'} (t) = 16 e^{2 t} + c_{1}

Solve

y^{'} (t) = 16 e^{2 t} + c_{1} : y = 8 e^{2 t} + c_{1} t + c_{2}

Apply initial conditions:

y = 8 e^{2 t} - 8 t - 5

y = 8 e^{2 t} - 8 t - 5

\int_{a}^{b} \frac{1}{b - a}

16 y^{^{''}} + 24 y^{^{'}} + 9 y = 0, y (2) = 4, y^{^{'}} (- 2) = - 3

- 3 cos (- 3 x)

\frac{d}{d x} (x - 1 - ln (x))

x^{2} (1 - y) \frac{d y}{d x} + y^{2} (1 + x) = 0

What is the solution for y^{''}=32e^{2t},y(0)=3,y^'(0)=8 ?
The solution for y^{''}=32e^{2t},y(0)=3,y^'(0)=8 is y=8e^{2t}-8t-5