What is the solution for (x^2+1)y^'+6x(y-1)=0,y(0)=8 ?

The solution for (x^2+1)y^'+6x(y-1)=0,y(0)=8 is y= 7/((x^2+1)^3)+1

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(x^2+1)y^'+6x(y-1)=0,y(0)=8

(x^{2} + 1) y^{^{'}} + 6 x (y - 1) = 0, y (0) = 8

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

Solution steps

(x^{2} + 1) y^{'} (x) + 6 x (y - 1) = 0

Solve separable ODE:

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

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

\int (ln (y))^{2} d y

a re a x 4 - x^{2}, 0

s im pl i f y x + y

a re a y = \frac{1}{x}, x = 2, y = - 1

\int_{- \infty}^{- 1} x e^{x} d x

What is the solution for (x^2+1)y^'+6x(y-1)=0,y(0)=8 ?
The solution for (x^2+1)y^'+6x(y-1)=0,y(0)=8 is y= 7/((x^2+1)^3)+1