## - Separable Differential Equations

The solution of this type of equation can be achieved by separating the variables and integrating both sides of the equation with respect to the relative variables.

LESSON 1
Solve the differential equations
$$\displaystyle {dy\over dx}={5x^2-3\over y}$$
SOLUTION
$$\displaystyle {dy\over dx}={5x^2-3\over y}$$
• Separate variables.
$$y \:dy=(5x^2-3)\:dx$$
• Integrate both sides with respect to the required variable.
$$∫y\:dy=∫(5x^2-3)\:dx$$
$$\displaystyle {y^2\over2}={5x^3\over3}-3x+c$$
• Simplify
$$\displaystyle y^2={10x^3\over3}-6x+c$$

LESSON 2
Solve the differential equation
$$\displaystyle {dy\over dx}=\cos ⁡x \tan ⁡y$$
SOLUTION
$$\displaystyle {dy\over dx}=\cos ⁡x \tan ⁡y$$
• Separate variables.
$$\displaystyle {1\over \tan ⁡y}\:dy=\cos ⁡x \:dx$$
• Integrate both sides with respect to the required variable.
$$\displaystyle ∫{\cos⁡ y\over \sin ⁡y}\:dy=∫\cos ⁡x\:dx$$
$$\ln⁡(\sin ⁡y)=\sin ⁡x+c$$
• Take e of both sides and simplify.
$$e^{\ln⁡ (\sin ⁡y)}=e^{\sin⁡ x+c}$$
$$\sin ⁡y=e^c e^{\sin ⁡x}$$
$$\sin ⁡y=Ae^{\sin ⁡x}$$ where $$A=e^c$$
These solutions are called general solutions of the differential equation because the value of the constant is unknown.

LESSON 3
Find the particular solution of the differential equation
$$\displaystyle x^2 {dy\over dx}=\csc ⁡y$$
when $$\displaystyle x=4, y={π\over3}$$.
SOLUTION
$$\displaystyle x^2 {dy\over dx}=\csc ⁡y; y={π\over3}$$ when $$x=4$$
• Separate variables
$$\displaystyle {1\over \csc ⁡y} dy={1\over x^2} dx$$
• Integrate both sides.
$$\displaystyle ∫{1\over \csc ⁡y}\:dy=∫x^{-2}\:dx$$
$$\int \sin ⁡y\:dy=\int x^{-2} dx$$
$$\displaystyle -\cos ⁡y=-{1\over x}+c$$
• Use the given values to determine the value of $$c$$.
$$\displaystyle -\cos \left({π\over3} \right)=-{1\over 4}+c$$
$$\displaystyle -{1\over 2}=-{1\over4}+c$$
$$\displaystyle -{1\over4}=c$$
• Substitute determined value of $$c$$.
$$\displaystyle \cos ⁡y={1\over x}+{1\over 4}$$