Differential Equations of First Order But Not of First Degree Differential Equations Solvable For X
Let f(x,y,p) = 0 be the given differential equation ……………………(1)
If equation (1) cannot be split up into rational and linear factors and (1) is of first degree in x, then (1) can be solved for x.
(1) can be expressed in the form x = F(y,p) ……………………….(2)
Differentiating (2) w.r.t. y gives an equation of the form \(\frac{1}{p}=g\left(y, p, \frac{d p}{d y}\right)\) …………………(3)
Since (3) is an equation in two variables p and y , it can be solved.
∴ The solution of (3) is \(\phi(y, p, c)=0\) …………………(4).
Eliminating p from (1) and (4), general solution of (1) is \(\psi(x, y, c)=0\).
Note 1. If it is not possible to eliminate p, then the values of x and y in terms of p in the form \(x=f_1(p, c)\) and \(y=f_2(p, c)\) together give the general solution.
2. This method is especially useful for equations y being absent.
Differential Equations of First Order But Not of First Degree Solved Problems
Example. 1. Solve \(y^2 \log y=x p y+p^2\)
Solution.
Given equation is \(y^2 \log y=x p y+p^2\) …………………………(1)
Since x is the first degree in (1), it can be solved for x. \(x=\frac{y \log y}{p}-\frac{p}{y}\) ………………………(2)
Differentiating (2) w.r.t. y: \(\frac{1}{p}=(1+\log y) \frac{1}{p}-\frac{y \log y}{p^2} \frac{d p}{d y}-\frac{1}{y} \frac{d p}{d y}+\frac{p}{y^2}\)
⇒ \(\frac{1}{p}=\frac{1}{p}+\frac{1}{\mathrm{p}} \log y+\frac{p}{y^2}-\left(\frac{y \log y}{p^2}+\frac{1}{y}\right) \frac{d p}{d y}\)
⇒ \(\frac{p}{y}\left(\frac{y \log y}{p^2}+\frac{1}{y}\right)-\left(\frac{y \log y}{p^2}+\frac{1}{y}\right) \frac{d p}{d y}=0\)
⇒ \(\left(\frac{y \log y}{p^2}+\frac{1}{y}\right)\left(\frac{p}{\mathrm{y}}-\frac{d p}{d y}\right)=0 \Rightarrow \frac{y \log y}{p^2}+\frac{1}{y}=0\) ………………………(3)
∴ \(\frac{p}{y}-\frac{d p}{d y}=0\) ……………………….(4)
(3) is discarded as it gives a singular solution
Solving(4) : \(\frac{d p}{d y}=\frac{p}{y} \Rightarrow \frac{d p}{p}=\frac{d y}{y} \Rightarrow \int \frac{d p}{p}=\int \frac{d y}{y}+\log c\)
⇒ \(\log p=\log y+\log c \Rightarrow \log p=\log c y \Rightarrow p=c y\) …………………………(5)
Eliminating p from (1) and (5): \(y^2 \log y=c x y^2+c^2 y^2 \Rightarrow \log y=c x+c^2\)
∴ The general solution of (1) is \(\log y=c x+c^2\)
Differential Equations Of First Order But Not First Degree
Example. 2. Solve \(x=y+p^2\)
Solution.
Given equation is \(x=y+p^2\) ………………………….(1)
Differentiating (1) w.r.t. \(y \Rightarrow \frac{d x}{d y}=1+2 p \frac{d p}{d y} \Rightarrow \frac{d p}{d y}=\frac{1-p}{2 p^2}\)
Separating the variables : \(d y=\frac{-2 p^2}{p-1} d p \Rightarrow \int d y=\int \frac{-2 p^2}{p-1} d p+c\)
⇒ \(y=-2 \int\left(p+1+\frac{1}{p-1}\right) d p+c \Rightarrow-2\left[\frac{1}{2} p^2+p+\log (p-1)\right]+c\) ………………………..(2)
Substituting the value of y from (2) in (1), we get: \(x=c-2[p+\log (p-1)]\) ………………………(3)
which shows that it is not possible to eliminate p from (1) and (2).
∴ The general solution of(1) is \(x=c-2[p+\log (p-1)], y=-p^2-2 p-2 \log (p-1)+c\)
Differential Equations Solvable For x Explained
Example. 3. Solve \(a y p^2+(2 x-b) p-y=0\) where p = dy / dx, a and h are real numbers.
Solution.
Given equation is \(a y p^2+(2 x-b) p-y=0\)
Since x is of first degree in (1), it can be solved for x. \(\Rightarrow 2 x=\frac{y}{p}-a y p+b\)
Differentiating w.r.t.y: \(\frac{2}{p}=\frac{1}{p} \cdot 1-\frac{y}{p^2} \frac{d p}{d y}-a\left(p+y \frac{d p}{d y}\right)\)
⇒ \(\frac{1}{p}+\frac{y}{p^2} \frac{d p}{d y}+a\left(p+y \frac{d p}{d y}\right)=0\)
⇒ \(\frac{1}{p^2}\left(p+y \frac{d p}{d y}\right)+a\left(p+y \frac{d p}{d y}\right)=0 \Rightarrow\left(p+y \frac{d p}{d y}\right)\left(\frac{1}{p^2}+a\right)=0\)
∴ \(p+y \frac{d p}{d y}=0\) ……………………..(2)
∴ \(\frac{1}{p^2}+a=0\) …………………….(3)
(3) is discarded as it gives a singular solution.
Solving (2): \(p+y \frac{d p}{d y}=0 \Rightarrow \frac{d p}{p}+\frac{d y}{y}=0\) (separating variables)
Integrating : \(\int \frac{d p}{p}+\int \frac{d y}{y}=\log c \Rightarrow \log p+\log y=\log c\)
⇒ \(\log p y=\log c \Rightarrow p y=c \Rightarrow p=c / y\)………………….(4)
By eliminating p from (1) and (4) => \(\text { ay }\left(\frac{c^2}{y^2}\right)+(2 x-b) \frac{c}{y}-y=0\)
∴ The general solution of (1) is \(a c^2+(2 x-b) c-y^2=0\)
Examples Of Differential Equations Solvable For x
Example. 4. Solve \(a p^2+p y-x=0\)
Solution.
Given equation is \(x=y p+a p^2\) ……………………..(1)
Differentiating (1) w.r.t.y \(\Rightarrow \frac{d x}{d y}=p+y \frac{d p}{d y}+2 a p \frac{d p}{d y}\)
⇒ \(\frac{1}{p}-p=(y+2 a p) \frac{d p}{d y} \Rightarrow \frac{d y}{d p}=\frac{p y+2 a p^2}{1-p^2}\)
⇒ \(\frac{d y}{d p}-\frac{p}{1-p^2} y=\frac{2 a p^2}{1-p^2}\) which is linear in y and p ……………………….(2)
∴ \(\text { I.F. }=\exp \left(\int \frac{-p}{1-p^2} d p\right)=\exp \left[\frac{1}{2} \log \left(1-p^2\right)\right]=\sqrt{1-p^2}\)
The general solution of (2) is \(y \sqrt{1-p^2}=2 a \int \frac{p^2}{1-p^2}, \sqrt{1-p^2} d p+c\)
⇒ \(y \sqrt{1-p^2}=2 a \int \frac{1-\left(1-p^2\right)}{\sqrt{1-p^2}} d p+c=2 a \int \frac{d p}{\sqrt{1-p^2}}-2 a \int \sqrt{1-p^2} d p+c\)
⇒ \(y \sqrt{1-p^2}=2 a \sin ^{-1} p-2 a\left[\frac{p}{2} \sqrt{1-p^2}+\frac{1}{2} \sin ^{-1} p\right]+c\)
= \(a \sin ^{-1} p-a p \sqrt{1-p^2}+c \Rightarrow y=\frac{a \sin ^{-1} p+c}{\sqrt{1-p^2}}-a p\) ……………………….(3)
By eliminating y from (1) and (3) \(\Rightarrow x=\frac{p}{\sqrt{1-p^2}}\left(a \sin ^{-1} p+c\right)\) …………………….(4)
∴ (3) and (4) together form the general solution of (1).
First-Order Equations Not Of First Degree With Solutions
Example. 5. Solve \(\)
Solution.
Given \(p^3-4 x y p+8 y^2=0 \Rightarrow 4 x=\frac{p^2}{y}+\frac{8 y}{p}\) …………………………(1)
Differentiating (1) w.r.t.y \(\Rightarrow \frac{4}{p}=\frac{2 p}{y} \frac{d p}{d y}-\frac{p^2}{y^2}+\frac{8}{p}-\frac{8 y}{p^2} \frac{d p}{d y}\)
⇒ \(\frac{p^2}{y^2}-\frac{4}{p}=\left(\frac{2 p}{y}-\frac{8 y}{p^2}\right) \frac{d p}{d y} \Rightarrow \frac{p^3-4 y^2}{p y^2}=\frac{2\left(p^3-4 y^2\right)}{p^2 y} \frac{d p}{d y}\)
⇒ \(\left(p^3-4 y^2\right)\left(\frac{1}{y}-\frac{2}{p} \frac{d p}{d y}\right)=0 \Rightarrow p^3-4 y^2=0\) …………………….(2)
\(\frac{1}{y}-\frac{2}{p} \frac{d p}{d y}=0\) …………………….(3)
(2) is discarded as it gives a singular solution.
Solving (3): \(\frac{d y}{y}-\frac{2}{p} d p=0 \Rightarrow \int \frac{d y}{y}-2 \int \frac{d p}{p}=-\log c\)
⇒ \(\log y-2 \log p=-\log c \Rightarrow \log y+\log c=\log p^2\)
⇒ \(\log (c y)=\log p^2 \Rightarrow p^2=y c\) ………………………(4)
Eliminating p from (1) and (4): general solution of (1) is
8 \(y^2=p\left(4 x y-p^2\right)=p(4 x y-c y) \Rightarrow 64 y^4=p^2 y^2(4 x-c)^2\)
⇒ \(64 y^4=c y^3(4 x-c)^2 \Rightarrow 64 y=c(4 x-c)^2\)
Problems On First-Order Non-First-Degree Differential Equations
Example. 6. Solve \(2 p x=2 \tan y+p^3 \cos ^2 y\)
Solution.
Given equation is \(2 p x=2 \tan y+p^3 \cos ^2 y\) ……………………..(1)
Since x is of first degree in (1), it can be solved for x:
∴ \(x=\frac{\tan y}{p}+\frac{p^2 \cos ^2 y}{2}\) ……………………..(2)
Differentiating (2) w.r.t.y:
⇒ \(\frac{1}{p}=\frac{1}{p} \sec ^2 y-\frac{1}{p^2} \tan y \frac{d p}{d y}+p \cos ^2 y \frac{d p}{d y}-p^2 \sin y \cos y\)
⇒ \(\left[\frac{1}{p} \tan ^2 y-p^2 \sin y \cos y\right]+\left(p \cos ^2 y-\frac{\tan y}{p^2}\right) \frac{d p}{d y}=0\)
⇒ \(-p \tan y\left(p \cos ^2 y-\frac{\tan y}{\dot{p}^2}\right)+\left(p \cos ^2 y-\frac{\tan y}{p^2}\right) \frac{d p}{d y}=0\)
⇒ \(\left(p \cos ^2 y-\frac{\tan y}{p^2}\right)\left(\frac{d p}{d y}-p \tan y\right)=0\)
∴ \(p \cos ^2 y-\frac{\tan y}{p^2}=0\) ………………………(3)
∴ \(\frac{d p}{d y}-p \tan y=0\) ……………………..(4)
(3) is rejected as it gives a singular solution.
Solving (4): \(\frac{d p}{d y}=p \tan y \Rightarrow \frac{d p}{p}=\tan y d y\)
⇒ \(\int \frac{d p}{p}=\int \tan y d y+\log c \Rightarrow \log p=\log \sec y+\log c\)
⇒ \(\log p=\log (c \sec y) \Rightarrow p=c \sec y\) ………………………(5)
Eliminating p from (1) and (5) \(\Rightarrow 2 c x y=2 \tan y+c^3 \sec ^3 y \cos ^2 y \Rightarrow 2 c x=2 \sin y+c^3\)
∴ The general solution of (1) is \(2 c x=2 \sin y+c^3\).