Differential Equations of First Order But Not of First Degree Equations That Do Not Contain X(or Y)
If the equation is of form f(x,p) = 0 and is solvable for p, it will give \(\frac{d y}{d x}=\phi(x)\), which is integrable. But if it is solvable for x, then x = F(p).
If the equation is of form f(y,p) = 0 and is solvable for p, it will give \(\frac{d y}{d x}=\phi(y)\), which is integrable. But if it is solvable for y, then y = F(p).
Differential Equations of First Order But Not of First Degree Solved Problems
First-Order But Not First-Degree Equations Examples And Solutions
Example. 1. Solve \(x p^3=a+b p\)
Solution.
Given equation is \(x p^3=a+b p\) ………………………(1)
Solving (1) for x, we have : \(x=\frac{a}{p^3}+\frac{b}{p^2}\) …………………………(2)
Differentiating (2) w.r.t. \(y \Rightarrow \frac{d x}{d y}=-\frac{3 a}{p^4} \frac{d p}{d y}-\frac{2 b}{p^3} \frac{d p}{d y}\)
⇒ \(\frac{1}{p}=-\frac{1}{p}\left(\frac{3 a}{p^3}+\frac{2 b}{p^2}\right) \frac{d p}{d y} \Rightarrow 1=-\left(\frac{3 a}{p^3}+\frac{2 b}{p^2}\right) \frac{d p}{d y}\)
⇒ \(d y=-\left(\frac{3 a}{p^3}+\frac{2 b}{p^2}\right) d p\) (by separating variables)
Integrating : \(\int d y=-3 a \int \frac{1}{p^3} d p-2 b \int \frac{1}{p^2} d p+c \Rightarrow y=\frac{3 a}{2 p^2}+\frac{2 b}{p}+c\) …………………..(3)
It is not possible to eliminate p from (1) and (3):
∴ General solution of (1) is \(x=\frac{a}{p^3}+\frac{b}{p^2}\) and \(y=\frac{3 a}{2 p^2}+\frac{2 b}{p}+c\)
Differential Equations Solved Without X Or Y Variables
Example. 2. Solve \(x \sqrt{1+p^2}+p=a \sqrt{1+p^2}\)
Solution.
Given equation is \(x=a-\frac{p}{\sqrt{1+p^2}}\) …………………..(1)
Differentiating (1) w.r.t y: \(\frac{1}{p}-\frac{1}{\sqrt{1+p^2}} \frac{d p}{d y}+\frac{1}{2} p \frac{1}{\left(1+p^2\right)^{3 / 2}} 2 p \frac{d p}{d y}=0\)
⇒ \(\frac{1}{p}=-\frac{1}{\left(1+p^2\right)^{3 / 2}} \frac{d p}{d y} \Rightarrow d y=-\frac{p}{\left(1+p^2\right)^{3 / 2}} d p\)
⇒ \(\int d y=-\frac{1}{2} \int \frac{2 p}{\left(1+p^2\right)^{3 / 2}} d p-c \Rightarrow y+c=\frac{1}{\sqrt{1+p^2}} \Rightarrow(y+c)^2=\frac{1}{1+p^2}\) ………………….(2)
Eliminating p from (1) and (2)
From (1): \((x-a)^2=\frac{p^2}{1+p^2}=1-\frac{1}{1+p^2} \Rightarrow(x-a)^2=1-(y+c)^2\)
∴ The general solution of (1) is \((x-a)^2+(y+c)^2=1\).
Equations That Are First-Order But Differ From Standard Forms
Example. 3. Solve \(e^y=p^3+p\)
Solution.
Given equation is \(e^y=p^3+p\)
Differentiating (1) w.r.t. \(x \Rightarrow e^y \frac{d y}{d x}=3 p^2 \frac{d p}{d x}+\frac{d p}{d x}\)
⇒ \(p e^y=\left(3 p^2+1\right) \frac{d p}{d x} \Rightarrow p\left(p^3+p\right)=\left(3 p^2+1\right) \frac{d p}{d x}\) [from (1)]
Separating the variables: \(d x=\frac{3 p^2+1}{p^2\left(p^2+1\right)} d p\)
Integrating: \(\int d x=\int \frac{3 p^2+1}{p^2\left(p^2+1\right)} d p+c\)
⇒ \(x=\int\left(\frac{1}{p^2}+\frac{2}{p^2+1}\right) d p+c \Rightarrow x=-\frac{1}{p}+2 \text{Tan}^{-1} p+c \text {. }\)
It is not possible to eliminate p from (1) and (3).
∴ The G.S. of (1) is given by \(e^y=p^3+p\) and \(x=-\frac{1}{p}+2 \text{Tan}^{-1} p+c\).