What is the solution for (x^2+1)y^'+3x(y-1)=0,y(0)=8 ?

The solution for (x^2+1)y^'+3x(y-1)=0,y(0)=8 is y= 7/((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)y^'+3x(y-1)=0,y(0)=8

(x^{2} + 1) y^{^{'}} + 3 x (y - 1) = 0, y (0) = 8

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

Solution steps

(x^{2} + 1) y^{'} (x) + 3 x (y - 1) = 0

Solve separable ODE:

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

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

\int \frac{sin ( c x )}{cos ^{2} ( c x )} d x

\frac{14 x + 8 x ^{2}}{( 7 + 8 x ) ^{2}}

\frac{d}{d t} x + 7 x = 5 cos (2 t)

\int_{1}^{3} x 3^{x^{2}} d x

g (x) = 27 x^{2} - x^{3}

What is the solution for (x^2+1)y^'+3x(y-1)=0,y(0)=8 ?
The solution for (x^2+1)y^'+3x(y-1)=0,y(0)=8 is y= 7/((x^2+1)^{3/2)}+1