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sergiy2304 [10]
4 years ago
8

Solve the differential equation 3ydx+(xy+5x)dy=0

Mathematics
2 answers:
drek231 [11]4 years ago
8 0

Answer:

-3㏑x+C=y+5㏑y

Step-by-step explanation:

Given differential equation is separable,

3ydx +(xy+5x)dy= 0

3ydx=-(xy+5x)dy

taking x as common,

3ydx=-(y+5)xdy

-3/xdx=(y+5)/ydy

-3/xdx=(1+5/y)dy

integrating both sides of equation,

∫-3/xdx=∫(1+5/y)dy

-3㏑x+C=y+5㏑y which is the solution of given ordinary differential equation.C is the constant of integration.


Brrunno [24]4 years ago
4 0

The ODE is separable:

3y\,\mathrm dx+(xy+5x)\,\mathrm dy=0\implies x(y+5)\,\mathrm dy=-3y\,\mathrm dx\implies\dfrac{y+5}y\mathrm dy=-\dfrac3x\,\mathrm dx

Integrating both sides gives

\displaystyle\int\left(1+\frac5y\right)\,\mathrm dy=-3\int\frac{\mathrm dx}x\implies y+5\ln|y|=-3\ln|x|+C

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