What is the solution for (x^2+1)y^'+7x(y-1)=0 ?

The solution for (x^2+1)y^'+7x(y-1)=0 is y=(c_{1})/((x^2+1)^{7/2)}+1

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(x^2+1)y^'+7x(y-1)=0

(x^{2} + 1) y^{^{'}} + 7 x (y - 1) = 0

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

Solution steps

(x^{2} + 1) y^{'} (x) + 7 x (y - 1) = 0

Solve separable ODE:

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

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

(x - 2) y^{^{'}} - y = 2 (x - 2)^{3}, y (3) = 5

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

x^{2} + 16, at x = 3

\int \frac{1}{3} x^{3} 9 - x^{2} d x

\frac{d}{d x} (2 x^{2} + 5)

What is the solution for (x^2+1)y^'+7x(y-1)=0 ?
The solution for (x^2+1)y^'+7x(y-1)=0 is y=(c_{1})/((x^2+1)^{7/2)}+1