What is the solution for (d^2y)/(dx^2)=3e^{-x},y(0)=1,y^'(0)=4 ?

The solution for (d^2y)/(dx^2)=3e^{-x},y(0)=1,y^'(0)=4 is y=3e^{-x}+7x-2

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(d^2y)/(dx^2)=3e^{-x},y(0)=1,y^'(0)=4

\frac{d ^{2} y}{d x ^{2}} = 3 e^{- x}, y (0) = 1, y^{^{'}} (0) = 4

y = 3 e^{- x} + 7 x - 2

Solution steps

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

Solve

y^{''} (x) = 3 e^{- x} : y^{'} (x) = - 3 e^{- x} + c_{1}

Solve

y^{'} (x) = - 3 e^{- x} + c_{1} : y = 3 e^{- x} + c_{1} x + c_{2}

Apply initial conditions:

y = 3 e^{- x} + 7 x - 2

y = 3 e^{- x} + 7 x - 2

\int 1 \frac{1}{x ^{2}} d x

(\frac{( x + 3 )}{x - 1})^{2}

y^{^{'}} - 2 x = 4 x^{3}, y (0) = 1

7 x ln (\frac{16 x}{16 x + 1})

\frac{d y}{d x} = 2 x + 3 y

What is the solution for (d^2y)/(dx^2)=3e^{-x},y(0)=1,y^'(0)=4 ?
The solution for (d^2y)/(dx^2)=3e^{-x},y(0)=1,y^'(0)=4 is y=3e^{-x}+7x-2