What is the solution for (x^2+1)y^'+3x(y-1)=0,y(0)=5 ?

The solution for (x^2+1)y^'+3x(y-1)=0,y(0)=5 is y= 4/((x^2+1)^{3/2)}+1

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(x^2+1)y^'+3x(y-1)=0,y(0)=5

(x^{2} + 1) y^{^{'}} + 3 x (y - 1) = 0, y (0) = 5

y = \frac{4}{( x ^{2} + 1 ) ^{\frac{3}{2}}} + 1

Solution steps

(x^{2} + 1) y^{'} (x) + 3 x (y - 1) = 0

Solve separable ODE:

y = \frac{4}{( x ^{2} + 1 ) ^{\frac{3}{2}}} + 1

y = \frac{4}{( x ^{2} + 1 ) ^{\frac{3}{2}}} + 1

\int_{0}^{\infty} e^{- x^{2}} d x

\int \frac{- 5}{( x ^{2} + 5 ) ^{\frac{3}{2}}} d x

\frac{d}{d t} (\frac{6}{5} - t^{2})

\frac{x ^{2}}{8}, 2 x^{2}, 2, 0

n = 1 \sum \infty n! x^{n}

What is the solution for (x^2+1)y^'+3x(y-1)=0,y(0)=5 ?
The solution for (x^2+1)y^'+3x(y-1)=0,y(0)=5 is y= 4/((x^2+1)^{3/2)}+1