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

(x^{2} - 1) (x^{2} - 4), x = 0, x = 5

x 4 - x^{2}, 0

x + y

\frac{d}{d x} (4 x (x^{2} + 1))

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