What is the solution for (x^2+1)y^'+5xy=5x,y(0)=6 ?

The solution for (x^2+1)y^'+5xy=5x,y(0)=6 is y= 5/((x^2+1)^{5/2)}+1

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(x^2+1)y^'+5xy=5x,y(0)=6

(x^{2} + 1) y^{^{'}} + 5 x y = 5 x, y (0) = 6

y = \frac{5}{( x ^{2} + 1 ) ^{\frac{5}{2}}} + 1

Solution steps

(x^{2} + 1) y^{'} (x) + 5 x y = 5 x

Solve separable ODE:

y = \frac{5}{( x ^{2} + 1 ) ^{\frac{5}{2}}} + 1

y = \frac{5}{( x ^{2} + 1 ) ^{\frac{5}{2}}} + 1

\frac{d y}{d x} = 4 + y - 4 x + 6

2 x y \frac{d y}{d x} + x^{2} + y^{2} = 0

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

f (x) = 7 ln (3)

\int sec (x) \cdot tan^{2} (x) d x

What is the solution for (x^2+1)y^'+5xy=5x,y(0)=6 ?
The solution for (x^2+1)y^'+5xy=5x,y(0)=6 is y= 5/((x^2+1)^{5/2)}+1