What is the solution for (x^2+1)y^'+7x(y-1)=0,y(0)=4 ?

The solution for (x^2+1)y^'+7x(y-1)=0,y(0)=4 is y= 3/((x^2+1)^{7/2)}+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x^2+1)y^'+7x(y-1)=0,y(0)=4

(x^{2} + 1) y^{^{'}} + 7 x (y - 1) = 0, y (0) = 4

y = \frac{3}{( x ^{2} + 1 ) ^{\frac{7}{2}}} + 1

Solution steps

(x^{2} + 1) y^{'} (x) + 7 x (y - 1) = 0

Solve separable ODE:

y = \frac{3}{( x ^{2} + 1 ) ^{\frac{7}{2}}} + 1

y = \frac{3}{( x ^{2} + 1 ) ^{\frac{7}{2}}} + 1

x \to \infty lim (cos (0))

sin (x) + \frac{5}{8} cot (x)

x \to - 6 lim (x^{2} - 1)

n = 1 \sum \infty \frac{1}{e ^{2 n}}

(12 - 20 y + e^{- 5 x}) d x - 4 d y = 0

What is the solution for (x^2+1)y^'+7x(y-1)=0,y(0)=4 ?
The solution for (x^2+1)y^'+7x(y-1)=0,y(0)=4 is y= 3/((x^2+1)^{7/2)}+1