What is the solution for (x^2+1)y^'+4x(y-1)=0,y(0)=5 ?

The solution for (x^2+1)y^'+4x(y-1)=0,y(0)=5 is y= 4/((x^2+1)^2)+1

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(x^2+1)y^'+4x(y-1)=0,y(0)=5

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

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

Solution steps

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

Solve separable ODE:

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

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

\int x 16 - 25 x^{2} d x

\int_{0}^{\frac{π}{2}} 3 cos (x) d x

f (x) = (3 x^{3} + 2 x) (x - 2) (x + 4)

\int \frac{1}{( 4 x ^{2} - 25 ) ^{\frac{3}{2}}} d x

\int - 4 x^{2} - 12 x - 5 d x

What is the solution for (x^2+1)y^'+4x(y-1)=0,y(0)=5 ?
The solution for (x^2+1)y^'+4x(y-1)=0,y(0)=5 is y= 4/((x^2+1)^2)+1