MATH 129 lesson 10-12

 

    First Order but higher degree differential equation:

A differential equation involving first order but degree higher than one is called first order but higher degree differential equation.

    Example: $P^n+A_1 P^{n-1}+A_2 P^{n-2}+\ldots +A_{n-a} P+ A_n$

is a first-order and n-th degree differential equation, where  $p=\frac{d y}{d x}$ and $A_1, A_2, \ldots , A_n$

are functions of x,y

Convenient of solution and various types of form, The equation is solved by five methods,

1. Solvable for p
2. Solvable for y 
3. solvable for x
4. Clairauit's equation
5. Lagrange's equation

    Solvable for p:

If the equation $P^n+A_1 P^{n-1}+A_2 P^{n-2}+\ldots +A_{n-a} P+ A_n=0$ can be expressed into n-factors of the first order and first degree $(p-\alpha_1) (p-\alpha_2)(p-\alpha_3)\ldots (p-\alpha_n)=0$ then equating to zero each factor, n-equation of the first order and first degree are obtained, Then the solutions are obtained by integration.

    

    Question: Solve,  $p^2+2xp-3x^2=0$

    Solution: Given equation is, 

$$\begin{align*} & p^2+2xp-3x^2=0 \tag{1}\label{eq:1} \\ or, ~& p^2+3xp-xp-3x^2=0\\ or, ~& p(p+3x)-x(p+3x)=0 \\ or, ~& (p-x)(p + 3x)= 0\\ ~& \\ \therefore & ~ p - x = 0 \tag{2}\label{eq:2} \\ & ~ p + 3x = 0 \tag{3}\label{eq:3} \\ \end{align*}$$

From (2) we get, $$\begin{align*} & \frac{dy}{dx} - x = 0\\ & dy - x dx = 0 \end{align*}$$

Integrating this we get, $$\begin{align*} & y-\frac{x ^ 2 }{2}- \frac{c}{2}= 0 \\ or, ~ & 2y-x ^2 -c = 0 \end{align*}$$

From (3) we get, $$\begin{align*} & \frac{dy}{dx} +3x =0 \\ or,~ & dy + 3x~dx = 0 \end{align*}$$

Integrating this we get, $$\begin{align*} & y + \frac{3x2 ^2}{2} - \frac{c}{2}= 0\\ & 2y + 3x^2 - c = 0 \end{align*}$$

General solution of (1) is, $$\begin{align*} (2y - x^2 - c )(2y + 3x^2 - c ) = 0 \end{align*}$$