What is the solution for 8x^2y+y^'=-3x^2 ?

The solution for 8x^2y+y^'=-3x^2 is y=(e^{(-8x^3+c_{1})/3})/(8)-3/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

8x^2y+y^'=-3x^2

8 x^{2} y + y^{^{'}} = - 3 x^{2}

y = \frac{e ^{\frac{- 8 x ^{3} + c _{1}}{3}}}{8} - \frac{3}{8}

Solution steps

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

Solve separable ODE:

y = \frac{e ^{\frac{- 8 x ^{3} + c _{1}}{3}}}{8} - \frac{3}{8}

y = \frac{e ^{\frac{- 8 x ^{3} + c _{1}}{3}}}{8} - \frac{3}{8}

\frac{d}{d x} (x^{\frac{1}{2}} y)

r \cdot \frac{d q}{d r} + 3 q = 0

x \to 0 + lim (5 x + 6)

\int \frac{4 x sin ( y ^{2} )}{y ^{3}} d y

x \to - 1 lim (x + 1)

What is the solution for 8x^2y+y^'=-3x^2 ?
The solution for 8x^2y+y^'=-3x^2 is y=(e^{(-8x^3+c_{1})/3})/(8)-3/8