What is the solution for y^'+9xy=8x,y(0)=6 ?

The solution for y^'+9xy=8x,y(0)=6 is y=(46e^{-(9x^2)/2})/(9)+8/9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+9xy=8x,y(0)=6

y^{^{'}} + 9 x y = 8 x, y (0) = 6

y = \frac{46 e ^{- \frac{9 x ^{2}}{2}}}{9} + \frac{8}{9}

Solution steps

y^{'} (x) + 9 x y = 8 x

Solve separable ODE:

y = \frac{46 e ^{- \frac{9 x ^{2}}{2}}}{9} + \frac{8}{9}

y = \frac{46 e ^{- \frac{9 x ^{2}}{2}}}{9} + \frac{8}{9}

\frac{d y}{d x} = \frac{3 x ^{2} + 4 x + 2}{2 y + 1}

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

\frac{d y}{d t} = \frac{2}{100} - \frac{3 y}{100}

y \cdot d y - 2 x y^{2} \cdot d x = 0

\frac{y ^{^{'}}}{x} = - 6 y

What is the solution for y^'+9xy=8x,y(0)=6 ?
The solution for y^'+9xy=8x,y(0)=6 is y=(46e^{-(9x^2)/2})/(9)+8/9