2.1 Variable separable method

  Differential Equations

If the differential equation can be written in the form of $$f(x)dx = g(y)dy$$ means the variables $x$ are separated with $dx$ and variables $y$ are separated with $dy$ and solution of the equation is obtained by integrating both the sides separately i.e., $$\int {f(x)dx} = \int {g(y)dy} + C$$ where $C$ is an arbitrary constant.

Question 4. Solve the differential equation $$(1 + x)ydx = (y - 1)xdy$$

Solution: As it is clear from the given differential equation, we can easily separate the variables of $x$ and $y$ and apply variable separable method to solve the differential equation.

The equation can be written as

$$ \left( {\frac{{1 + x}}{x}} \right)dx = \left( {\frac{{y - 1}}{y}} \right)dy$$Integrating both sides$$ {\left( {\frac{1}{x} + 1} \right)} dx = \int {\left( {1 - \frac{1}{y}} \right)} dy $$ $$ \ln x + x = y - \ln y + c$$where $c$ is the constant $$ln x + \ln y = y - x + c $$$$ xy = y - x + c $$ $$ xy = {e^{y - x + c}} = {e^{y - x}}.{e^c} $$ $$ xy = C{e^{y - x}}$$where $C$ is another constant equals to $e^c$

The above solution of a differential equation is a general solution.

Question 5. Solve $${e^{\frac{{dy}}{{dx}}}} = x + 1$$ given that when $x=0$, $y=3$.

Solution: $$ {e^{\frac{{dy}}{{dx}}}} = x + 1$$taking ln both sides$$ \Rightarrow \frac{{dy}}{{dx}} = \ln (x + 1) $$ $$ \Rightarrow dy = \ln (x + 1)dx$$Integrating both sides $$ \Rightarrow \int {dy} = \int {\ln (x + 1)dx}$$Integrating by parts$$ y = x\ln (x + 1) - \int {\frac{x}{{x + 1}}dx} $$ $$ y = x\ln (x + 1) - x + \ln (x + 1) + C $$

This is a general solution. From the condition given in the question, put $x=0$ and $y=3$ in the general solution, we get,

$$3=0-0+0+C$$$$C=3$$

Therefore, the required particular solution is $$y = (x + 1)\ln (x + 1) - x + 3$$