What is the solution for (x^2+1)y^'+5x(y-1)=0,y(0)=9 ?

The solution for (x^2+1)y^'+5x(y-1)=0,y(0)=9 is y= 8/((x^2+1)^{5/2)}+1

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(x^2+1)y^'+5x(y-1)=0,y(0)=9

(x^{2} + 1) y^{^{'}} + 5 x (y - 1) = 0, y (0) = 9

y = \frac{8}{( x ^{2} + 1 ) ^{\frac{5}{2}}} + 1

Solution steps

(x^{2} + 1) y^{'} (x) + 5 x (y - 1) = 0

Solve separable ODE:

y = \frac{8}{( x ^{2} + 1 ) ^{\frac{5}{2}}} + 1

y = \frac{8}{( x ^{2} + 1 ) ^{\frac{5}{2}}} + 1

n = 0 \sum \infty q^{n}

\int \frac{1 + 8 v}{- 16 v ^{2}} d v

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

\int \frac{( x - 2 )}{4 x - x ^{2}} d x

\frac{d}{d x} (\frac{x - 5}{x + 5})

What is the solution for (x^2+1)y^'+5x(y-1)=0,y(0)=9 ?
The solution for (x^2+1)y^'+5x(y-1)=0,y(0)=9 is y= 8/((x^2+1)^{5/2)}+1