What is the solution for f^{''}(x)=4,f^'(2)=10,f(2)=15 ?

The solution for f^{''}(x)=4,f^'(2)=10,f(2)=15 is f(x)=2x^2+2x+3

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f^{''}(x)=4,f^'(2)=10,f(2)=15

f^{^{''}} (x) = 4, f^{^{'}} (2) = 10, f (2) = 15

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

Solution steps

f^{''} (x) = 4

Solve

f^{''} (x) = 4 : f^{'} (x) = 4 x + c_{1}

Solve

f^{'} (x) = 4 x + c_{1} : f (x) = 2 x^{2} + c_{1} x + c_{2}

Apply initial conditions:

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

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

\frac{\partial}{\partial x} (r t)

x^{2} + x y + 2 y^{2} = 8, (- 2, - 1)

\int 4 x - x x d x

\frac{d}{d x} (\frac{9}{4 - x})

f (x) = 3 x^{3}, at x = 3

What is the solution for f^{''}(x)=4,f^'(2)=10,f(2)=15 ?
The solution for f^{''}(x)=4,f^'(2)=10,f(2)=15 is f(x)=2x^2+2x+3