What is the solution for vyln(v)dy-y(vdy+ydv)=0 ?

The solution for vyln(v)dy-y(vdy+ydv)=0 is y(v)=e^{c_{1}}ln(v)-e^{c_{1}}

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vyln(v)dy-y(vdy+ydv)=0

v y ln (v) d y - y (v d y + y d v) = 0

y (v) = e^{c_{1}} ln (v) - e^{c_{1}}

Solution steps

v y (v) ln (v) d y (v) - y (v) (v d y (v) + y (v) d v) = 0

Solve separable ODE:

y (v) = e^{c_{1}} ln (v) - e^{c_{1}}

y (v) = e^{c_{1}} ln (v) - e^{c_{1}}

y^{^{'}} = 0.01398 y, y (0) = 150

\frac{d y}{d x} = (- 2 x + 1) y, y (0) = 6.8

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

3 x y^{2} y^{^{'}} = y^{3}

\frac{d N}{d t} = r N (t) (1 - \frac{N}{2})

What is the solution for vyln(v)dy-y(vdy+ydv)=0 ?
The solution for vyln(v)dy-y(vdy+ydv)=0 is y(v)=e^{c_{1}}ln(v)-e^{c_{1}}