What is the solution for (dy)/(dx)+(3xy^3)=0,y(0)=1 ?

The solution for (dy)/(dx)+(3xy^3)=0,y(0)=1 is y= 1/(sqrt(3x^2+1))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dy)/(dx)+(3xy^3)=0,y(0)=1

\frac{d y}{d x} + (3 x y^{3}) = 0, y (0) = 1

y = \frac{1}{3 x ^{2} + 1}

Solution steps

\frac{d y}{d x} + (3 x y^{3}) = 0

Solve separable ODE:

y = \frac{1}{3 x ^{2} + 1}

y = \frac{1}{3 x ^{2} + 1}

\frac{d}{d x} (\frac{3}{6 x + 3})

\int_{0}^{5} \frac{1}{x ^{\frac{21}{22}}} d x

\int \frac{sin ( 2 x )}{42 + cos ^{2} ( x )} d x

\int \frac{x ^{3} - 4 x + 1}{x ^{2} - 3 x + 2} d x

\frac{d}{d x} (\frac{( 2 x ^{2} - 1 ) ^{3}}{( 5 x - 8 ) ^{2}})

What is the solution for (dy)/(dx)+(3xy^3)=0,y(0)=1 ?
The solution for (dy)/(dx)+(3xy^3)=0,y(0)=1 is y= 1/(sqrt(3x^2+1))