Differential Equations of First Order But Not of First Degree Differential Equations Solvable For Y:
Let f(x, y, p) = 0 be the giiven differential equation ………………..(1)
If (1) cannot be resolved into two rational and linear factors and (1) is of first degree in y , then it can be solved for y .
(1) can be expressed in the form y = F(x,p) …………………….(2)
Differentiation of (2) w.r.t.x. gives an equation of the form \(p=g\left(x, p, \frac{d p}{d x}\right)\) ……………………(3)
Since (3) is an equation in two variables p and x, it can be solved.
The solution of (1) is ψ(x,y,c) = 0 ……………….(4)
Note 1. If it is not possible to eliminate P from (1) and (4), the general solution of (1) is given in the form (1) Φ(x, p, c) = 0, f(x,y,p) = 0 or \(x=f_1(p, c), y=f_2(p, c)\)
This is regarded as a parametric form of the required solution where p is regarded as a parameter.
2. This method is especially useful for equations in which x is absent.
Differential Equations of First Order But Not of First Degree Solved Problems
Example. 1. Solve \(y=2 p x-p^2\)
Solution.
Given equation is \(y=2 p x-p^2\) ……………………(1)
Differentiating (1) w.r.t.x => \(p=2 p+2 x \frac{d p}{d x}-2 p \frac{d p}{d x}\)
⇒ \(2(p-x) \frac{d p}{d x}=p \Rightarrow p \frac{d x}{d p}+2 x=2 p \Rightarrow \frac{d x}{d p}+\frac{2 x}{p}=2\) …………………….(2)
(2) is a linear equation in x.
Then \(\text { I.F. }=\exp \left[\int \frac{2}{p} d p\right]=\exp (2 \log p)=\exp \left(\log p^2\right)=p^2\)
∴ The general solution of (2) is x \((\mathrm{I} . \mathrm{F})=\int \mathrm{Q}(\mathrm{I} . \mathrm{F}) d p+\mathrm{c}\) where Q = 2.
⇒ \(x p^2=\int 2 p^2 d p+c=\frac{2 p^3}{3}+c \Rightarrow x=\frac{2}{3} p+\frac{c}{p^2}\) ………………………(3)
It is not possible to eliminate p from (1) and (3)
∴ The general solution of (1) is given by two equations:
x = \((2 p / 3)+\left(c / p^2\right) \text { and } y=2 p\left(\frac{2 p}{3}+\frac{c}{p^2}\right)-p^2=\frac{p^2}{3}+\frac{2 c}{p}\)
Differential Equations Of First Order But Not First Degree Solvable For y
Example. 2. Solve \(y=x p^2+p\)
Solution.
Given equation is \(y=x p^2+p\) …………………….(1)
Differentiating w.r.t. x => \(p=p^2+2 x p \frac{d p}{d x}+\frac{d p}{d x}\)
⇒ \(\left(p^2-p\right) \frac{d x}{d p}+2 x p+1=0 \Rightarrow \frac{d x}{d p}+\frac{2}{p-1} x=-\frac{1}{p(p-1)}\) ………………………..(2)
(2) is a linear equation in x where \(\mathrm{P}=\frac{2}{p-1}, \quad \mathrm{Q}=\frac{-1}{p(p-1)}\)
⇒ \(\int \mathrm{P} d p=\int \frac{2}{p-1} d p=2 \log (p-1)=\log (p-1)^2\)
∴ \(\text { I.F. }=e^{\log (p-1)^2}\)
∴ The general solution of (2) is \(x(\mathrm{I} . \mathrm{F})=\int \mathrm{Q}(\mathrm{I} . \mathrm{F}) d p+c \Rightarrow x(p-1)^2=\int \frac{-1}{p(p-1)}(p-1)^2 d p+c\)
⇒ \(x(p-1)^2=c-\int \frac{p-1}{p} d p=c-p+\log p\) ……………………..(3)
It is not possible to eliminate 7 from (1) and (3).
∴ The G.S. of (1) is given by \(x=(c-p+\log p)(p-1)^{-2} \text { and } y=x p^2+p\).
Examples Of Differential Equations Solvable For y
Example. 3. Solve \(y+p x=p^2 x^4\)
Solution.
Given equation is \(y+p x=p^2 x^4\) ………………………(1)
But (1) is of first degree in y and hence (1) can be solved for y.
Differentiating (1) w.r.t. x, we get \(p+p+x \frac{d p}{d x}=2 p x^4 \frac{d p}{d x}+4 p^2 x^3 \Rightarrow 2 p+\left(x-2 p x^4\right) \frac{d p}{d x}-4 p^2 x^3=0\)
⇒ \(x\left(1-2 x^3 p\right) \frac{d p}{d x}+2 p\left(1-2 x^3 p\right)=0 \Rightarrow\left(1-2 x^3 p\right)\left(x \frac{d p}{d x}+2 p\right)=0\)
⇒ \(1-2 x^3 p=0\) ………………………..(2)
x\(\frac{d p}{d x}+2 p=0\) ………………………..(3)
(2) is discarded as it gives a singular solution
Solving(3): \(x \frac{d p}{d x}+2 p=0 \Rightarrow \frac{d p}{p}+\frac{2 d x}{x}=0\)
Integrating: \(\int \frac{d p}{p}+2 \int \frac{d x}{x}=\log c \Rightarrow \log p+2 \log x=\log c\)
⇒ \(\log p+\log x^2=\log c \Rightarrow \log p x^2=\log c \Rightarrow p x^2=c \Rightarrow p=c / x^2\)
Eliminating p from (1) and (4), the general solution of (1) is \(y+(c / x)=\left(c^2 / x^4\right) x^4 \Rightarrow y+(c / x)=c^2\)
Note 1. From (2) : \(p=1 /\left(2 x^3\right)\) ………………….(5)
Eliminating p from (1) and (5)
⇒ \(y+\left(1 / 2 x^2\right)=1 /\left(4 / x^2\right) \Rightarrow y+1 /\left(4 x^2\right)=0 \Rightarrow 4 x^2 y+1=0\). which is called a singular solution since it does not contain an arbitrary constant.
2. While finding the G.S. the factor not involving \(\frac{d p}{d x}\) is discarded.
First-Order Equations Not Of First Degree Solvable For y Examples
Example. 4. Solve \(y=2 x p+x^2 p^4\)
Solution.
Given equation is \(y=2 x p+x^2 p^4\) ………………………..(1)
Differentiating (1) w.r.t. x, we get: \(\frac{d y}{d x}=2 p+2 x \frac{d p}{d x}+2 x p^4+4 x^2 p^3 \frac{d p}{d x}\)
⇒ \(p=2 p+2 x p^4+2 x\left(1+2 x p^3\right) \frac{d p}{d x}\)
⇒ \(p\left(1+2 x p^3\right)+2 x\left(1+2 x p^3\right) \frac{d p}{d x}=0 \Rightarrow\left(1+2 x p^3\right)\left(p+2 x \frac{d p}{d x}\right)=0\)
⇒ \(1+2 x p^3=0\) …………………….(2)
p\(+2 x \frac{d p}{d x}=0\) ……………………(3)
(2) is discarded as it gives a singular solution.
Solving(3): \(p+2 x \frac{d p}{d x}=0 \Rightarrow \frac{2 d p}{p}+\frac{d x}{x}=0\)
Integrating: \(\Rightarrow 2 \int \frac{d p}{p}+\int \frac{d x}{x}=\log c \Rightarrow 2 \log p+\log x=\log c\)
⇒ \(\log p^2 x=\log c \Rightarrow p^2 x=c \Rightarrow p^2=c / x\) …………………….(4)
Eliminating p from (1) and (4) \(\Rightarrow\left(y-x^2 p^4\right)^2=4 x^2 p^2\)
∴ The general solution of (1) is \(\left[y-\left(x^2 c^2 / x^2\right)\right]^2=4 x^2(c / x) \Rightarrow\left(y-c^2\right)^2=4 c x\).
Examples Of Solvable For y Differential Equations
Example. 5. Solve \(x p^2-2 y p+x=0\)
Solution.
Given equation is \(x p^2-2 y p+x=0\) ………………………..(1)
Differentiating (1) w.r.t. \(x \Rightarrow p^2+2 p x \frac{d p}{d x}-2 p \frac{d y}{d x}-2 y \frac{d p}{d x}+1=0\)
⇒ \(p^2+2(p x-y) \frac{d p}{d x}-2 p^2+1=0 \Rightarrow\left(1-p^2\right)+2(p x-y) \frac{d p}{d x}=0\) ………………………..(2)
Eliminating y from (1) and (2) : \(\left(1-p^2\right)+2\left(p x-\frac{x p^2+x}{2 p}\right) \frac{d p}{d x}=0\)
⇒ \(\left(1-p^2\right)+\frac{2\left(2 p^2 x-x p^2-x\right)}{2 p} \frac{d p}{d x}=0 \Rightarrow p\left(1-p^2\right)+\left(p^2-1\right) x \frac{d p}{d x}=0\)
⇒ \(\left(1-p^2\right)\left(p-x \frac{d p}{d x}\right)=0 \Rightarrow 1-p^2=0\) …………………….(3)
p\(-x \frac{d p}{d x}=0\) ………………………(4)
(3) is discarded as it does not contain \(\) and it gives the singular solution.
Solving (4) : \(p=x \frac{d p}{d x} \Rightarrow \frac{d x}{x}=\frac{d p}{p}\)
Integrating: \(\int \frac{d p}{p}=\int \frac{d x}{x}+\log c \Rightarrow \log p=\log x+\log c \Rightarrow \log p=\log c x \Rightarrow p=c x\) ……………………(5)
Eliminating p from (1) and (5): G.S.. of(1) is \(c^2 x^3-2 c x y+x=0 \Rightarrow c^2 x^2-2 c y+1=0\).
Solved Examples Of Differential Equations Solvable For y
Example. 6. Solve \(y+x p \log p=(2+3 \log p) p^3\)
Solution.
Given equation is \(y=-x p \log p+(2+3 \log p) p^3\) ………………………..(1)
Solving for y: Differentiating (1) w.r.t. x
⇒ \(p=-p \log p-x(\log p+1) \frac{d p}{d x}+\frac{3}{p} \cdot p^3 \frac{d p}{d x}+(2+3 \log p) 3 p^2 \frac{d p}{d x}\)
⇒ \(p(1+\log p)=-x(\log p+1) \frac{d p}{d x}+9 p^2(1+\log p) \frac{d p}{d x}\)
⇒ \((1+\log p)\left[\left(-x+9 p^2\right) \frac{d p}{d x}-p\right]=0\) ……………………(2)
⇒ \(1+\log p=0\) …………………..(3)
p = \(\left(-x+9 p^2\right) \frac{d p}{d x}\) …………………..(4)
(3) is discarded as it gives the singular solution.
Solving (4): \(p \frac{d x}{d p}+x=9 p^2 \Rightarrow \frac{d x}{d p}+\frac{1}{p} x=9 p\) ……………………(5)
(5) is a linear equation in x.
∴ I.F = \(\exp \left(\int \frac{1}{p} d p\right)=e^{\log p}=p\)
The general solution of(5) is \(x_p=\int 9 p \cdot p d p+c=3 p^2+c\) ……………………..(6)
Eliminating x from (1) and (6): \(y=-\left(3 p^2+\frac{c}{p}\right) p \log p+(2+3 \log p) p^3\)
⇒ \(y=-3 p^3 \log p-c \log p+2 p^3+3 p^3 \log p=2 p^3-c \log p\) …………………….(7)
It is not possible to eliminate p from (6) and (7).
∴ The general solution of (1) is \(x=3 p^2+c p^{-1} \text { and } y=2 p^3-c \log p\)
Example. 7. Solve \(x-y p=a p^2\)
Solution.
Given equation is \(x-y p=a p^2 \Rightarrow y=\frac{x}{p}-a p\) ………………………..(1)
Differentiating (1) w.r.t x => \(\frac{d y}{d x}=p=\frac{1}{p}-\frac{x}{p^2} \frac{d p}{d x}-a \frac{d p}{d x}\)
⇒ \(\left(a p^2+x\right) \frac{d p}{d x}=p\left(1-p^2\right) \Rightarrow p\left(1-p^2\right) \frac{d x}{d p}-x=a p^2\)
⇒ \(\frac{d x}{d p}-\frac{1}{p\left(1-p^2\right)} x=\frac{a p}{1-p^2}\) is a linear equation in x ………………………….(2)
where P = \(-\frac{1}{p\left(1-p^2\right)}\) and \(\mathrm{Q}=\frac{a p}{1-p^2}\)
⇒ \(\int \mathrm{P} d p=\int \frac{1}{p\left(1-p^2\right)} d p=\int\left[\frac{1}{p}+\frac{1}{2(1-p)}-\frac{1}{2(1+p)}\right] d p\)
= \(\log p-\frac{1}{2} \log (1-p)-\frac{1}{2} \log (1+p)=\log \left(\frac{p}{\sqrt{1-p^2}}\right)\)
∴ \(\text { I.F. }=\exp \left[\int \mathrm{P} d p\right]=\exp \left[-\log \frac{p}{\sqrt{1-p^2}}\right]=\exp \left[\log \frac{\sqrt{1-p^2}}{p}\right]=\frac{\sqrt{1-p^2}}{p}\)
The general solution of (2) is \(\frac{x \sqrt{1-p^2}}{p}=\int \frac{a p}{1-p^2} \cdot \frac{\sqrt{1-p^2}}{p} d p+c\)
⇒ \(\frac{x \sqrt{1-p^2}}{p}=a \int \frac{d p}{\sqrt{1-p^2}}+c=a \sin ^{-1} p+c \Rightarrow x=\frac{p}{\sqrt{1-p^2}}\left(a \sin ^{-1} p+c\right)\) …………………….(3)
It is not possible to eliminate p from (1) and (3)
∴ The general solution of (1) is given by the two equations
y = \(x p^2+p \text { and } x=\frac{p}{\sqrt{1-p^2}}\left(a \sin ^{-1} p+c\right)\)
Example 8. Solve : \(x^2+p^2 x=y p\)
Solution.
Given equation is \(x^2+p^2 x=y p\) ……………………….(1)
Solving for y: \(y=p x+\left(x^2 / p\right)\)
Differentiating w.r.t.x : \(\frac{d y}{d x}=p=p+x \frac{d p}{d x}+\frac{p(2 x)-x^2(d p / d x)}{p^2}\)
⇒ \(x \frac{d p}{d x}+\frac{2 p x-x^2(d p / d x)}{p^2}=0 \Rightarrow p^2 x \frac{d p}{d x}+2 p x-x^2 \frac{d p}{d x}=0\)
⇒ \(x\left(p^2-x\right) \frac{d p}{d x}=-2 p x \Rightarrow\left(p^2-x\right) \frac{d p}{d x}=-2 p \Rightarrow\left(x-p^2\right) \frac{d p}{d x}=2 p \Rightarrow 2 p \frac{d x}{d p}-x=-p^2\)
⇒ \(\frac{d x}{d p}-\frac{1}{2 p} x=-\frac{p}{2}\). This is a linear equation in x. ………………………..(2)
Now I.F. = \(e^{\int-(1 / 2 p) d p}=e^{-(1 / 2) \log p}=e^{\log p^{-1 / 2}}=p^{-1 / 2}\)
∴ The general solution of (2) is \(x \cdot p^{-1 / 2}=\int-\frac{p}{2} \cdot p^{-1 / 2} d p+c\)
⇒ \(\frac{x}{\sqrt{p}}=-\frac{1}{2} \int p^{1 / 2} d p+c \Rightarrow \frac{x}{\sqrt{p}}=-\frac{1}{2} \frac{p^{3 / 2}}{3 / 2}+c=-\frac{1}{3} p^{3 / 2}+c\)
⇒ \(x=c \sqrt{p}-(1 / 3) p^2\) …………………………(3)
Substituting x from (3) in (1):
y = \(p\left(c \sqrt{p}-(1 / 3) p^2\right)+\frac{\left[c \sqrt{p}-(1 / 3) p^2\right]^2}{p}\) …………………………(4)
(3) and (4) together is the solution of (1)