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

The solution for (x^2+1)((dy)/(dx))+3x(y-1)=0,y(0)=6 is y= 5/((x^2+1)^{3/2)}+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x^2+1)((dy)/(dx))+3x(y-1)=0,y(0)=6

(x^{2} + 1) (\frac{d y}{d x}) + 3 x (y - 1) = 0, y (0) = 6

y = \frac{5}{( x ^{2} + 1 ) ^{\frac{3}{2}}} + 1

Solution steps

(x^{2} + 1) (\frac{d y}{d x}) + 3 x (y - 1) = 0

Solve separable ODE:

y = \frac{5}{( x ^{2} + 1 ) ^{\frac{3}{2}}} + 1

y = \frac{5}{( x ^{2} + 1 ) ^{\frac{3}{2}}} + 1

\frac{1}{3 ( 4 - x ^{3} ) ^{4}}

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

\int_{1}^{e} ln (16 x) d x

x \to 1 lim (\frac{4 - 3 x}{1 - x})

\int_{1}^{1000} \frac{17}{x ^{3}} d x

What is the solution for (x^2+1)((dy)/(dx))+3x(y-1)=0,y(0)=6 ?
The solution for (x^2+1)((dy)/(dx))+3x(y-1)=0,y(0)=6 is y= 5/((x^2+1)^{3/2)}+1