What is the solution for x^2y^'-2xy=3y^4,y(1)= 1/2 ?

The solution for x^2y^'-2xy=3y^4,y(1)= 1/2 is y=(\sqrt[3]{5}x^2)/(\sqrt[3]{-9x^5+49)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2y^'-2xy=3y^4,y(1)= 1/2

x^{2} y^{^{'}} - 2 x y = 3 y^{4}, y (1) = \frac{1}{2}

Solution steps

x^{2} y^{'} (x) - 2 x y = 3 y^{4}

Solve Bernoulli ODE:

\int \frac{1}{e ^{x} + e ^{2 x}} d x

\int \frac{tan ^{3} ( ln ( x ) )}{7 x} d x

8 e^{3 x} + e^{- x^{2}}

\int \frac{x ^{5}}{( 1 - x ^{6} ) ^{7}} d x

\frac{d}{d y} ((y - 12)^{\frac{3}{2}})

What is the solution for x^2y^'-2xy=3y^4,y(1)= 1/2 ?
The solution for x^2y^'-2xy=3y^4,y(1)= 1/2 is y=(\sqrt[3]{5}x^2)/(\sqrt[3]{-9x^5+49)}