What is the solution for f^{''}(x)=e^x,f^'(0)=2,f(0)=5 ?

The solution for f^{''}(x)=e^x,f^'(0)=2,f(0)=5 is f(x)=e^x+x+4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

f^{''}(x)=e^x,f^'(0)=2,f(0)=5

f^{^{''}} (x) = e^{x}, f^{^{'}} (0) = 2, f (0) = 5

f (x) = e^{x} + x + 4

Solution steps

f^{''} (x) = e^{x}

Solve

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

Solve

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

Apply initial conditions:

f (x) = e^{x} + x + 4

f (x) = e^{x} + x + 4

\int (4 x^{3} - 9 x^{2} + 16 x) x^{4} - 3 x^{3} + 8 x^{2} d x

x \to 0 lim (\frac{4 + x - 4 - x}{4 x})

x \to - \frac{π}{2} lim (- tan (0.5 x))

\int_{0}^{\frac{π}{2}} x^{2} sin (x) d x

x \to \infty lim (\frac{2 x ^{2}}{x + 3})

What is the solution for f^{''}(x)=e^x,f^'(0)=2,f(0)=5 ?
The solution for f^{''}(x)=e^x,f^'(0)=2,f(0)=5 is f(x)=e^x+x+4