What is the solution for (x^2+1)y^'=-4x(y-1),y(0)=8 ?

The solution for (x^2+1)y^'=-4x(y-1),y(0)=8 is y= 7/((x^2+1)^2)+1

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(x^2+1)y^'=-4x(y-1),y(0)=8

(x^{2} + 1) y^{^{'}} = - 4 x (y - 1), y (0) = 8

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

Solution steps

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

Solve separable ODE:

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

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

\int 4 e^{5} d x

x \to 0 lim (\frac{e ^{\frac{1}{x}}}{x})

t^{2} y^{^{'}} - t = 1 + y + t y

\int \frac{x + 5}{x ^{2} + 4 x + 8} d x

f (x) = e^{3 x}

What is the solution for (x^2+1)y^'=-4x(y-1),y(0)=8 ?
The solution for (x^2+1)y^'=-4x(y-1),y(0)=8 is y= 7/((x^2+1)^2)+1