What is the solution for (yp+2y)(pdy+ydp)-(2yp+4y)dy=0 ?

The solution for (yp+2y)(pdy+ydp)-(2yp+4y)dy=0 is p=(c_{1})/y+2

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(yp+2y)(pdy+ydp)-(2yp+4y)dy=0

(y p + 2 y) (p d y + y d p) - (2 y p + 4 y) d y = 0

p = \frac{c _{1}}{y} + 2

Solution steps

(y p + 2 y) (p d y + y d p) - (2 y p + 4 y) d y = 0

Solve separable ODE:

p = \frac{c _{1}}{y} + 2

p = \frac{c _{1}}{y} + 2

cos^{2} (13 t)

y = ln (5)

\frac{d y}{d x} = 5 cos^{3} (3 x + 6)

\int \frac{x ^{3}}{x ^{2} - 16} d x

\int_{1}^{2} \frac{12}{x ^{2}} - 1 d x

What is the solution for (yp+2y)(pdy+ydp)-(2yp+4y)dy=0 ?
The solution for (yp+2y)(pdy+ydp)-(2yp+4y)dy=0 is p=(c_{1})/y+2