Differential Equations of First Order and First Degree Solved Example Problems Exercise 2.6
Example. 1. Solve \(\left(y^2+2 x^2 y\right) d x+\left(2 x^3-x y\right) d y=0\)
Solution.
Given equation is \(\left(y^2+2 x^2 y\right) d x+\left(2 x^3-x y\right) d y=0\)
⇒ \(y\left(y+2 x^2\right) d x+x\left(2 x^2-y\right) d y=0\) . Let \(x^h y^k\) be the integrating factor.
Multiplying the given equation by \(x^h y^k\).
⇒ \(\left(x^h y^{k+2}+2 x^{h+2} y^{k+1}\right) d x+\left(2 x^{h+3} y^k-x^{h+1} y^{k+1}\right) d y=0\) …………………..(1)
where \(\mathrm{M}=x^h y^{k+2}+2 x^{h+2} y^{k+1}, \mathrm{~N}=2 x^{h+3} y^k-x^{h+1} y^{k+1}\)
If (1) exact , then \(\frac{\partial \mathrm{M}}{\partial y}=\frac{\partial \mathrm{N}}{\partial x}\)
⇒ \((k+2) x^h y^{k+1}+2(k+1) x^{h+2} y^k=-(h+1) x^h y^{k+1}+2(h+3) x^{h+2} y^k\)
Equating the coefficients of \(x^h y^{k+1} \text { and } x^{h+2} y^k\)
we get \(k+2=-(h+1), k+1=h+3 \Rightarrow h+k+3=0 ; h-k+2=0\)
Solving these equations: h = -5/2,k = -1/2.
∴ Integrating factor is [latexx^{-5 / 2} y^{-1 / 2}][/latex].
Multiplying the given equation with this I. F. we get \(\left(x^{-5 / 2} y^{3 / 2}+2 x^{-1 / 2} y^{1 / 2}\right) d x+\left(2 x^{1 / 2} y^{-1 / 2}-x^{-3 / 2} y^{1 / 2}\right) d y=0\) …………………….(2)
(2) is an exact equation where \(\mathrm{M}_1=x^{-5 / 2} y^{3 / 2}+2 x^{-1 / 2} y^{1 / 2}, \mathrm{~N}_1=2 x^{1 / 2} y^{-1 / 2}-x^{-3 / 2} y^{1 / 2}\) and \(\frac{\partial \mathrm{M}_1}{\partial y}=\frac{\partial \mathrm{N}_1}{\partial x}\) now solving (2).
⇒ \(\int \mathrm{M}_1 d x=\int\left(x^{-5 / 2} y^{3 / 2}+2 x^{-1 / 2} y^{1 / 2}\right) d x\) treating y as constant.
= \(y^{3 / 2} x^{-3 / 2}(-2 / 3)+4 x^{1 / 2} y^{1 / 2} \text { and } \int \mathrm{N}_1 d y\) = integrating the terms not having x = 0.
The required solution is \(\int \mathrm{M}_1 d x+\int \mathrm{N}_1 d y=c \Rightarrow-(2 / 3) x^{-3 / 2} y^{3 / 2}+4 x^{1 / 2} y^{1 / 2}=c\)
Differential Equations Of First Order And First Degree Exercise 2.6
Example. 2. Solve \((2 y d x+3 x d y)+2 x y(3 y d x+4 x d y)=0\)
Solution.
The given equation can be written as \(\left(2 y+6 x y^2\right) d x+\left(3 x+8 x^2 y\right) d y=0\) …………………(1)
Let \(x^h y^k\) be the integrating factor. Multiplying (1) by \(x^h y^k\) :
⇒ \(\left(2 x^h y^{k+1}+6 x^{h+1} y^{2+k}\right) d x+\left(3 x^{h+1} y^k+8 x^{h+2} y^{k+1}\right) d y=0\) ………………….(2)
(2) is an exact equation if \(\frac{\partial \mathrm{M}}{\partial y}=\frac{\partial \mathrm{N}}{\partial x}\) where
M\(=2 x^h y^{k+1}+6 x^{h+1} y^{2+k}, \mathrm{~N}=3 x^{h+1} y^k+8 x^{h+2} y^{k+1}\)
⇒ \(2(k+1) y^k x^h+6(k+2) y^{k+1} x^{h+1}=3(h+1) x^h y^k+8(h+2) x^{h+1} y^{k+1}\)
⇒ \(2 k+2=3 h+3,6 k+12=8 h+16 \Rightarrow 3 h-2 k+1=0,8 h-6 k+4=0\)
Solving these equations we get h=1,k=2.
∴ I.F. = xy²
Multiplying (1) with \(x y^2 \Rightarrow\left(2 x y^3+6 x^2 y^4\right) d x+\left(3 x^2 y^2+8 x^3 y^3\right) d y=0\)
where \(\dot{\mathrm{M}}_1=2 x y^3+6 x^2 y^4 \text { and } \mathrm{N}_1=3 x^2 y^2+8 x^3 y^3\)
⇒ \(\int \mathrm{M}_1 d x=\int\left(2 x y^3+6 x^2 y^4\right) d x=2\left(x^2 / 2\right) y^3+6\left(x^3 / 3\right) y^4\) treating y constant.
⇒ \(\int \mathrm{N}_1 d y\) = integrating the terms of N, not having x = 0.
∴ Solution of(1) is \(x^2 y^3+2 x^3 y^4+c=0 \Rightarrow x^2 y^3+2 x^3 y^4=c\)
Homogeneous Equations Solved Problems Exercise 2.6
3. Solve: \(\left(2 x^2 y-3 y^4\right) d x+\left(3 x^3+2 x y^3\right) d y=0\) ………………….(1)
Solution:
Given
\(\left(2 x^2 y-3 y^4\right) d x+\left(3 x^3+2 x y^3\right) d y=0\) ………………….(1)
Let \(x^a y^b\) be the integrating factor. multiplying (1) with this factor, we get :
⇒ \(\left(2 x^{a+2}+y^{b+1}-3 x^a y^{b+4}\right) d x+\left(3 x^{a+3} y^b+2 x^{a+1} y^{b+3}\right) d y=0\)
where \(\mathrm{M}=2 x^{a+2} y^{b+1}-3 x^a y^{b+4}, \quad \mathrm{~N}=3 x^{a+3} y^b+2 x^{a+1} y^{b+3}\) …………………..(2)
For (2) to be exact: \(\frac{\partial \mathrm{M}}{\partial y}=\frac{\partial \mathrm{N}}{\partial x}\)
⇒ \(2(b+1) x^{a+2} y^b-3(b+4) x^a y^{b+3}=3(a+3) x^{a+2} y^b+2(a+1) x^a y^{b+3}\)
⇒ \(2(b+1)=3(a+3), 2(a+1)=-3(b+4) \Rightarrow 3 a-2 b+7=0, \quad 2 a+3 b+14=0\)
Solving for a and b: a=-49/13,b=-28/13
∴ I. F = \(x^{-49 / 13} y^{-28 / 13}\)
Multiplying (1) by \(x^{-49 / 13} y^{-28 / 13}\) we get: \(\left(2 x^2 y \cdot x^{-49 / 13} y^{-28 / 13}-3 y^4 x^{-49 / 13} y^{-28 / 13}\right) d x\)
+\(\left(3 x^3 \cdot x^{-49 / 13} y^{-28 / 13}-2 x y^3 x^{-49 / 13} y^{-28 / 15}\right) d y=0\)
⇒ \(\left(2 x^{-23 / 13} \cdot y^{-15 / 13}-3 x^{-49 / 13} \cdot y^{24 / 13}\right) d x+\left(3 x^{-10 / 13} y^{-28 / 13}+2 x^{-36 / 13} y^{11 / 13}\right) d y=0\)
where \(\mathrm{M}_1=2 x^{-23 / 13} y^{-15 / 13}-3 x^{-49 / 13} y^{24 / 13}\) and \(\mathrm{N}_1=3 x^{-10 / 13} y^{-28 / 13}+2 x^{-36 / 13} y^{11 / 13}\)
⇒ \(\int \mathrm{M}_1 d x=2 y^{-15 / 13} \frac{x^{-10 / 13}}{-10 / 13}-3 y^{24 / 13} \frac{x^{-36 / 13}}{-36 / 13}=5 x^{-36 / 13} y^{24 / 13}-12 x^{-10 / 13} y^{-15 / 13}, \mathrm{~N}_1 d y=0\)
G S. is \(5 x^{-36 / 13} y^{24 / 13}-12 x^{-10 / 13} y^{-15 / 13}=c\)
Methods To Find Integrating Factors For Exercise 2.6
Example.4. Solve \(x(4 y d x+2 x d y)+y^3(3 y d x+5 x d y)=0\)
Solution.
Given equation is \(\left(4 x y+3 y^4\right) d x+\left(2 x^2+5 x y^3\right) d y=0\) …………….(1)
verify \(\frac{\partial \mathrm{M}}{\partial y} \neq \frac{\partial \mathrm{N}}{\partial x}\) => (1) is not an exact equation.
Let \(x^a y^b\) be an integrating factor. Multiplying (1) with I.F.
⇒ \(\left(4 x^{a+1} y^{b+1}+3 x^a y^{b+4}\right) d x+\left(2 x^{a+2} y^b+5 x^{a+1} y^{b+3}\right) d y=0\) …………………(2)
where \(\mathrm{M}=4 x^{a+1} y^{b+1}+3 x^a y^{b+4}, \mathrm{~N}=2 x^{a+2} y^b+5 x^{a+1} y^{b+3}\)
⇒ \(\frac{\partial \mathrm{M}}{\partial y}=4 x^{a+1}(b+1) y^b+3(b+4) x^a y^{b+3}, \frac{\partial \mathrm{N}}{\partial x}=2(a+2) x^{a+1} \cdot y^b+5(a+1) x^a y^{b+3}\)
for(2) to be an exact, \(\frac{\partial \mathrm{M}}{\partial y}=\frac{\partial \mathrm{N}}{\partial x}\)
⇒ \(4(b+1) x^{a+1} y^b+3(b+4) x^a y^{b+3}=2(a+2) x^{a+1} y^b+5(a+1) x^a y^{b+3}\)
⇒ \(4(b+1)=2(a+2), 3(b+4)=5(a+1) \Rightarrow a-2 b=0,5 a-3 b-7=0\)
Solving these equation: 10b – 36 = 7 ⇒ b =1 ∴ a= 2
Multiplying (1) by \(x^2 \cdot y\); we get: \(\left(4 x^3 y^2+3 x^2 y^5\right) d x+\left(2 x^4 y+5 x^3 y^4\right) d y=0\)
Here \(\mathrm{M}_1=4 x^3 y^2+3 x^2 y^5, \mathrm{~N}_1=2 x^4 \dot{y}+5 x^3 y^4\)
⇒ \(\int M_1 d x=\int\left(4 x^3 y^2+3 x^2 y^5\right) d x=\frac{4 x^4}{4} y^2+\frac{3 x^3}{3} y^5 . \int N_1 d y=0\)
∴ GS. is \(x^4 y^2+x^3 y^5=c\)