What is the solution for (dy)/(dx)=5x^4-sec^2(x) ?

The solution for (dy)/(dx)=5x^4-sec^2(x) is y=x^5-tan(x)+c_{1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dy)/(dx)=5x^4-sec^2(x)

\frac{d y}{d x} = 5 x^{4} - sec^{2} (x)

y = x^{5} - tan (x) + c_{1}

Solution steps

\frac{d y}{d x} = 5 x^{4} - sec^{2} (x)

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = 5 x^{4} - sec^{2} (x)

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

then

f (x) = \int g (x) d x

y = \int 5 x^{4} - sec^{2} (x) d x

\int 5 x^{4} - sec^{2} (x) d x = x^{5} - tan (x) + c_{1}

y = x^{5} - tan (x) + c_{1}

\int (8 x + 7) \cdot ln (x) d x

\frac{\partial}{\partial x} (2 y^{2} - 2 x y + 7 x - 5)

\int (x \cdot sec^{2} (x)) d x

\int x^{7} + 7 x + \frac{7}{x} + 7 d x

\int e^{x^{3} + 5} x^{2} d x

What is the solution for (dy)/(dx)=5x^4-sec^2(x) ?
The solution for (dy)/(dx)=5x^4-sec^2(x) is y=x^5-tan(x)+c_{1}