What is the solution for -(2dy)/(dx)+6x^2=0 ?

The solution for -(2dy)/(dx)+6x^2=0 is y=x^3+c_{1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

-(2dy)/(dx)+6x^2=0

- \frac{2 d y}{d x} + 6 x^{2} = 0

y = x^{3} + c_{1}

Solution steps

- \frac{2 d y}{d x} + 6 x^{2} = 0

Let

y

be the dependent variable. Divide by

d x

- \frac{2 \cdot \frac{d y}{d x}}{1} + 6 x^{2} = 0

Substitute

\frac{d y}{d x}

with

y^{'} (x)

- \frac{2 y ^{^{'}} ( x )}{1} + 6 x^{2} = 0

Isolate

y^{'} (x) : y^{'} (x) = 3 x^{2}

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

then

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

y = \int 3 x^{2} d x

\int 3 x^{2} d x = x^{3} + c_{1}

y = x^{3} + c_{1}

\int 6 x^{2} (2 x^{3} - 5)^{5} d x

\frac{d}{d x} (4 (1 - x)^{- 2})

\frac{d}{d x} (e^{- x})

\frac{\partial}{\partial x} (- 3 e^{3 y} sin (3 x))

x = y^{2}, x = y + 2 [0, 4]

What is the solution for -(2dy)/(dx)+6x^2=0 ?
The solution for -(2dy)/(dx)+6x^2=0 is y=x^3+c_{1}