What is the solution for y^{''}+12x=0,y(0)=-4,y^'(0)=5 ?

The solution for y^{''}+12x=0,y(0)=-4,y^'(0)=5 is y=-2x^3+5x-4

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y^{''}+12x=0,y(0)=-4,y^'(0)=5

y^{^{''}} + 12 x = 0, y (0) = - 4, y^{^{'}} (0) = 5

y = - 2 x^{3} + 5 x - 4

Solution steps

y^{''} (x) + 12 x = 0

Isolate

y^{''} (x) : y^{''} (x) = - 12 x

Solve

y^{''} (x) = - 12 x : y^{'} (x) = - 6 x^{2} + c_{1}

Solve

y^{'} (x) = - 6 x^{2} + c_{1} : y = - 2 x^{3} + c_{1} x + c_{2}

Apply initial conditions:

y = - 2 x^{3} + 5 x - 4

y = - 2 x^{3} + 5 x - 4

\frac{\partial}{\partial x} (\frac{6 y}{x + y})

\int e^{x} \cdot (1 - e^{x})^{- 1} d x

\frac{d}{d x} (2 x^{3} e^{x})

\frac{3}{x}

f (x) = x^{3} - 2 x + 1

What is the solution for y^{''}+12x=0,y(0)=-4,y^'(0)=5 ?
The solution for y^{''}+12x=0,y(0)=-4,y^'(0)=5 is y=-2x^3+5x-4