Question

Solve the following differential equation: (1− 5 y +x) dy/dx +y= 5/x −1 . C=

Answer 1

Answer:

$y(1+x)+x-5\ln xy=C$

Step-by-step explanation:

Consider the given differential equation is

$(1-\frac{5}{y}+x)\frac{dy}{dx}+y=\frac{5}{x}-1$

$(1-\frac{5}{y}+x)\frac{dy}{dx}=\frac{5}{x}-1-y$

$(1-\frac{5}{y}+x)dy=(\frac{5}{x}-1-y)dx$

Taking all variables on right sides.

$(1-\frac{5}{y}+x)dy-(\frac{5}{x}-1-y)dx=0$

$(-\frac{5}{x}+1+y)dx+(1-\frac{5}{y}+x)dy=0$

Let as assume,

$M=-\frac{5}{x}+1+y$ and $N=1-\frac{5}{y}+x$

Find partial derivatives $\frac{\partial M}{\partial y}$ and $\frac{\partial N}{\partial x}$

$\frac{\partial M}{\partial y}=1$ and $\frac{\partial N}{\partial x}=1$

Since $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$, therefore the given differential equation is exact.

The solution of the exact differential equation is

$\int Mdx+\int N(\text{without x)}dy=C$

$\int (-\frac{5}{x}+1+y)dx+\int (1-\frac{5}{y})dy=C$

$yx-5\ln x+x+y-5\ln y=C$

$y+x+xy-5\ln x-5\ln y=C$

$y(1+x)+x-5(\ln x+\ln y)=C$

$y(1+x)+x-5\ln xy=C$