Differential Equations

Separable Equations<!-- --> | Differential Equations

Separable Equations

Sharan Sajiv Menon - October 5th, 2021


A separable equation is a differential equation which can be written in the form y’=f(x)g(y)y’=f(x)g(y) or dydx=f(x)g(y)\frac{dy}{dx}=f(x)g(y)

Examples include:

  • y′=(x2−4)(3y+2)y'=(x^2-4)(3y+2)
  • y′=6x2+4xy'=6x^2+4x
  • y′=sec(y)+cot(y)y'=sec(y) + cot(y)

Separable equations are solved by separation of variables.

Separation of Variables

  • Check for any values of yy that make g(y)=0g(y)=0. Those are the constant solutions
  • Rewrite the equation so that the y-variables are on the dy side and the x variables are on the dx side
  • Integrate: ∫1g(y)dy=∫f(x)dx\int \frac{1}{g(y)}dy=\int f(x)dx
  • Solve the resulting equation for yy

Example 1: Solve the following differential equation: dydx=xy\frac{dy}{dx}=xy

To rewrite this equation, we can start by multiplying both sides by dxdx, to remove it from the left side.

dy=xydxdy=xydx

Now, we need to move the y-variable to a different side, to ensure complete separation of variables. We can do this by dividing both sides by yy.

dyy=xdx\frac{dy}{y}=xdx

Now, we can integrate.

∫dyy=∫xdx⟶ln(y)=x22+C\int \frac{dy}{y}=\int xdx \longrightarrow ln(y)=\frac{x^2}{2} + C

The general solution to a differential equation is in a form of y=y=, so we need to solve for y.

y=ex22+C=eCex22=Cex22y=e^{\frac{x^2}{2} + C}= e^{C}e^{\frac{x^2}{2}} = Ce^{\frac{x^2}{2}}

Note here that we can simply rewrite ece^c as CC, since ece^c is a constant. So, our general solution then becomes

y=Cex22y=Ce^{\frac{x^2}{2}}

Example 2: Solve the Differential Equation y’=(x2−4)(3y−2)y’=(x^2-4)(3y-2)

Solution: Use separation of variables to solve the ODE.

dydx=(x2−4)(3y−2)→dydx(3y−2)=x2−4→13y−2dy=(x2−4)dx\frac{dy}{dx}=(x^2-4)(3y-2)\rightarrow \frac{dy}{dx(3y-2)}=x^2-4 \rightarrow \frac{1}{3y-2}dy=(x^2-4)dx

Integrate the separated DE and solve for yy:

∫13y−2dy=∫(x2−4)dx⟶13ln∣3y−2∣=x33−4x+C1⟶ln∣3y−2∣=x3−12x+C1⟶3y−2=ex3−12x+C1→y=C1ex3−12x+23\int \frac{1}{3y-2}dy = \int (x^2-4)dx \longrightarrow \frac{1}{3}ln|3y-2|=\frac{x^3}{3}-4x +C_1 \longrightarrow ln|3y-2|=x^3-12x + C_1 \longrightarrow \\ 3y-2=e^{x^3-12x+C_1}\rightarrow y=\frac{C_1e^{x^3-12x}+2}{3}

Our final answer is y=C1ex3−12x+23y=\frac{C_1e^{x^3-12x}+2}{3}.

Note that we jumped directly from ex3−12x+C1e^{x^3-12x+C_1} to C1ex3−12xC_1e^{x^3-12x}. This is because ex3−12x+C1=e^{x^3-12x+C_1}=

Created by Sharan Sajiv Menon, © 2022