What is the solution for (x^2+1)(dy)/(dx)+3xy=3x,y(0)=2 ?

The solution for (x^2+1)(dy)/(dx)+3xy=3x,y(0)=2 is y= 1/((x^2+1)^{3/2)}+1

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(x^2+1)(dy)/(dx)+3xy=3x,y(0)=2

(x^{2} + 1) \frac{d y}{d x} + 3 x y = 3 x, y (0) = 2

y = \frac{1}{( x ^{2} + 1 ) ^{\frac{3}{2}}} + 1

Solution steps

(x^{2} + 1) \frac{d y}{d x} + 3 x y = 3 x

Solve separable ODE:

y = \frac{1}{( x ^{2} + 1 ) ^{\frac{3}{2}}} + 1

y = \frac{1}{( x ^{2} + 1 ) ^{\frac{3}{2}}} + 1

(x + 2) sin (y) + x cos (y) \frac{d y}{d x} = 0

\frac{d y}{d x} = - 600 y

f^{^{'}} (x) = 2 f (x)

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

y^{^{'}} = \frac{( ( x - 2 ) ^{2} )}{y}, y (0) = 1

What is the solution for (x^2+1)(dy)/(dx)+3xy=3x,y(0)=2 ?
The solution for (x^2+1)(dy)/(dx)+3xy=3x,y(0)=2 is y= 1/((x^2+1)^{3/2)}+1