What is the solution for (t^2+1)dy+3tydt=6tdt,y(0)=5 ?

The solution for (t^2+1)dy+3tydt=6tdt,y(0)=5 is y=(c_{1})/((t^2+1)^{3/2)}+2

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(t^2+1)dy+3tydt=6tdt,y(0)=5

(t^{2} + 1) d y + 3 t y d t = 6 t d t, y (0) = 5

y = \frac{c _{1}}{( t ^{2} + 1 ) ^{\frac{3}{2}}} + 2

Solution steps

(t^{2} + 1) d y + 3 t y d t = 6 t d t

Solve separable ODE:

y = \frac{c _{1}}{( t ^{2} + 1 ) ^{\frac{3}{2}}} + 2

y = \frac{c _{1}}{( t ^{2} + 1 ) ^{\frac{3}{2}}} + 2

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

v^{^{'}} = \frac{( v + \frac{4}{v} )}{t}

\frac{d y}{d x} 6 x y + 1 = 0

(x^{2} + 6)^{\frac{1}{2}} d x + \frac{1}{x csc ( 8 y )} d y = 0

\frac{d y}{d x} = 72 \cdot y \cdot x^{7}

What is the solution for (t^2+1)dy+3tydt=6tdt,y(0)=5 ?
The solution for (t^2+1)dy+3tydt=6tdt,y(0)=5 is y=(c_{1})/((t^2+1)^{3/2)}+2