Differential Equations Introduction Solved Problems
Reducible To Homogeneous Differential Equation Examples With Solutions
Example. 1: Solve : \((x+y-1) \frac{d y}{d x}=x-y+2\)
Solution.
Given equation is \(\frac{d y}{d x}=\frac{x-y+2}{x+y-1}\) ………………..(1)
where \(a_1=1, b_1=1, a_2=1, b_2=-1 \Rightarrow a_1 b_2-a_2 b_1=2 \neq 0\)
Put \(x=\mathrm{X}+h \text { and } y=\mathrm{Y}+k \text { in (1) } \Rightarrow \frac{d y}{d x}=\frac{d \mathrm{Y}}{d \mathrm{X}}\) ……………………….(2)
(1) and (2) => \(\frac{d \mathrm{Y}}{d \mathrm{X}}=\frac{\mathrm{X}-\mathrm{Y}+(h-k+2)}{\mathrm{X}+\mathrm{Y}+(h+k-1)}\)
Choose h and k such that h-k + 2 = 0 ……………….(4)
and h + k – 1 = 0 …………………..(5)
Solving (4) and (5): \(h=-\frac{1}{2}, k=\frac{3}{2}\)
Now (3), (4), (5) => \(\frac{d \mathrm{Y}}{d \mathrm{X}}=\frac{\mathrm{X}-\mathrm{Y}}{\mathrm{X}+\mathrm{Y}}\) ………………….(6)
∵ f(kX, kY) = f(X, Y), (6) is homogeneous equation.
Putting \(\mathrm{Y}=v \mathrm{X} \Rightarrow \frac{d \mathrm{Y}}{d \mathrm{X}}=v+\mathrm{X} \frac{d v}{d \mathrm{X}}\) in (6), we get \(v+\mathrm{X} \frac{d v}{d \mathrm{X}}=\frac{1-v}{1+v}\)
⇒ \(\mathrm{X} \frac{d v}{d \mathrm{X}}=\frac{1-v}{1+v}-v=\frac{-v^2-2 v+1}{1+v}\)
Separating the variable: \(\frac{1+v}{v^2+2 v-1} d v=-\frac{d \mathrm{X}}{\mathrm{X}} \Rightarrow \frac{1}{2} \int \frac{2 v+2}{v^2+2 v-1} d v=-\int \frac{d \mathrm{X}}{\mathrm{X}}+\frac{1}{2} \log c\)
⇒ \(\frac{1}{2} \log \left(v^2+2 v-1\right)=-\log \mathrm{X}+\frac{1}{2} \log c \Rightarrow \log \left(\frac{\mathrm{Y}^2}{\mathrm{X}^2}+\frac{2 \mathrm{Y}}{\mathrm{X}}-1\right)+2 \log \mathrm{X}=\log c\)
(∵ \(v=\frac{\mathrm{Y}}{\mathrm{X}}\))
⇒ \(\log \frac{\mathrm{Y}^2+2 \mathrm{XY}-\mathrm{X}^2}{\mathrm{X}^2}+\log \mathrm{X}^2=\log c \Rightarrow \log \left(\frac{\left(\mathrm{Y}^2+2 \mathrm{XY}-\mathrm{X}^2\right)}{\mathrm{X}^2} \cdot \mathrm{X}^2\right)=\log c\)
⇒ \(\mathrm{Y}^2+2 \mathrm{XY}-\mathrm{X}^2=c\)
Now substitute: \(X=x+\frac{1}{2}, Y=y-\frac{3}{2}\)
∴ The general solution of (1)[ is \(\left(y-\frac{3}{2}\right)^2+2\left(x+\frac{1}{2}\right)\left(y-\frac{3}{2}\right)-\left(x+\frac{1}{2}\right)^2=c\)
Aliter: The given equation can be written as
(x+2) d x-y d x=(y-1) d y+x d y
(x+2) d x=(y d x+x d y)+(y-1) d y[/latex]
⇒ \(\int(x+2) d x=\int d(x y)+\int(y-1) d y \Rightarrow\left(x^2 / 2\right)+2 x=x y+\left(y^2 / 2\right)-y+(c / 2)\)
∴ The solution is \(x^2-y^2-2 x y+4 x+2 y=c\)
Step-By-Step Examples Of Reducible To Homogeneous Differential Equations
Example. 2: Solve : \((4 x+3 y+1) d x+(3 x+2 y+1) d y=0\)
Solution.
Given equation is \(\frac{d y}{d x}=\frac{-(4 x+3 y+1)}{(3 x+2 y+1)}\) ……………(1)
where \(a_1=-4, b_1=-3, c_1=-1, a_2=3, b_2=2, c_2=1 \Rightarrow a_1 b_2-a_2 b_1 \neq 0\)
to solve (1): put \(x=\mathrm{X}+h, y=\mathrm{Y}+k \Rightarrow \frac{d y}{d x}=\frac{d \mathrm{Y}}{d \mathrm{X}}\) …………………(2)
(1) and (2) => \(\frac{d \mathrm{Y}}{d \mathrm{X}}=-\frac{4(\mathrm{X}+h)+3(\mathrm{Y}+k)+1}{3(\mathrm{X}+h)+2(\mathrm{Y}+k)+1}=\frac{-(4 \mathrm{X}+3 \mathrm{Y})-(4 h+3 k+1)}{(3 \mathrm{X}+2 \mathrm{Y})+3(h+2 k+1)}\) ……………………..(3)
Choose h and 4 such that 4h + 3k + 1 = 0 ………………………..(4)
and 3h + 2k + 1 = 0 …………………(5)
Solving (4) and (5) h = -1, k = 1.
Then (3), (4), (5) \(\Rightarrow \frac{d Y}{d X}=-\frac{4 X+3 Y}{3 X+2 Y}\) …………………..(6)
∵ f(hx, ky) = f(X, Y),, (6) is homogeneous equation.
Putting \(\mathrm{Y}=v \mathrm{X} \Rightarrow \frac{d \mathrm{Y}}{d \mathrm{X}}=v+\mathrm{X} \frac{d v}{d \mathrm{X}}\) in (6), we get
v\(+\mathrm{X} \frac{d v}{d \mathrm{X}}=-\frac{4+3 v}{3+2 v} \Rightarrow \mathrm{X} \frac{d v}{d \mathrm{X}}=\frac{-4-3 v}{3+2 v}-v=\frac{-2\left(v^2+3 v+2\right)}{3+2 v}\)
Separating the variables : \(\frac{2 v+3}{v^2+3 v+2} d v=-2 \frac{d \mathrm{X}}{\mathrm{X}}\)
Integrating: \(\int \frac{2 v+3}{v^2+3 v+2} d v=-2 \int \frac{d \mathrm{X}}{\mathrm{X}}+\log c \Rightarrow \log \left(v^2+3 v+2\right)=-2 \log \mathrm{X}+\log c\)
⇒ \(\log \left(v^2+3 v+2\right)+\log \mathrm{X}^2=\log c \Rightarrow \log \left(v^2+3 v+2\right) \mathrm{X}^2=\log c\)
(∵ \(V=\frac{\mathrm{Y}}{\mathrm{X}}\))
⇒ \(\left(v^2+3 v+2\right) \mathrm{X}^2=c \Rightarrow\left(\frac{\mathrm{Y}^2}{\mathrm{X}^2}+3 \frac{\mathrm{Y}}{\mathrm{X}}+2\right) \mathrm{X}^2=c\)
⇒ \(\mathrm{Y}^2+3 \mathrm{XY}+2 \mathrm{X}^2=c\)
Substituting X = x+1, Y = y -1, in this
The general solution of (1) is \((y-1)^2+3(x+1)(y-1)+2(x+1)^2=c\)
Aliter: Given equation can be written as –\((4 x+1) d x=3(y d x+x d y)+(2 y+1) d y \Rightarrow(-4 x-1)^{\prime} d x=3 d(x y)+(2 y+1) d y\)
Integrating : \(-2 x^2-x=3 x y+y^2+y\)
∴ G. S. is \(2 x^2+3 x y+y^2+x+y=c\)
Reducible To Homogeneous Differential Equation Solved Examples Tutorial
Example. 3 : Solve : \(\frac{d y}{d x}=\frac{(x+y-1)^2}{4(x-2)^2}\)
Solution.
Given equation is \(\frac{d y}{d x}=\left(\frac{x+y-1}{2 x-4}\right)^2\) …………………….(1)
where \(a_1=1, b_1=1, a_2=2, b_2=0 \Rightarrow a_1 b_2-a_2 b_1=0-2=-2 \neq 0\)
to solve (1): put \(x =\mathrm{X}+h, y=\mathrm{Y}+k \Rightarrow \frac{d y}{d x}=\frac{d \mathrm{Y}}{d \mathrm{X}}\) ……………………(2)
(1) and (2) => \(\frac{d \mathrm{Y}}{d \mathrm{X}}=\frac{(\mathrm{X}+h+\mathrm{Y}+k-1)^2}{4(\mathrm{X}+h-2)^2}=\frac{[(\mathrm{X}+\mathrm{Y})+(h+k-1)]^2}{4[\mathrm{X}+(h-2)]^2}\) …………………….(3)
Choose, h and k such that h + k – 1 = 0 ……………………….(4)
and h – 2 = 0 ………………………..(5)
Solving (4) and (5): h = 2, k = -1
∴ \(\mathrm{X}=x-2, \mathrm{Y}=y-(-1)=y+1\)
(3), (4), (p) => \(\) …………………(6)
∵ f(kX, kY) = f(X, Y), (6) is homogeneous equation
Putting \(\mathrm{Y}=v \mathrm{X} \Rightarrow \frac{d \mathrm{Y}}{d \mathrm{X}}=v+\mathrm{X} \frac{d v}{d \mathrm{X}}\) in (6), we get
v\(+x \frac{d v}{d \mathrm{X}}=\frac{(\mathrm{X}+v \mathrm{X})^2}{4 \mathrm{X}^2}=\frac{(1+v)^2}{4} \Rightarrow \mathrm{X} \frac{d v}{d \mathrm{X}}=\frac{(1+v)^2}{4}-v=\frac{(1+v)^2-4 v}{4}=\frac{(1-v)^2}{4}\)
Separating the variables : \(\frac{4}{(1-v)^2} d v=\frac{d \mathrm{X}}{\mathrm{X}} \Rightarrow 4 \int \frac{d v}{(1-v)^2}=\int \frac{d \mathrm{X}}{\mathrm{X}}+c\)
⇒ \(4 \cdot \frac{-1}{1-v} \cdot \frac{1}{-1}=\log \mathrm{X}+c \Rightarrow \frac{4}{1-(\mathrm{Y} / \mathrm{X})}=\log \mathrm{X}+c\)
(∵ \(v=\frac{\mathrm{Y}}{\mathrm{X}}\))
⇒ \(\frac{4 X}{X-Y}=\log X+c\) ……………………(7)
Substituting X = x-2, Y = y +1, in (7)
∴ The general solution of (1) is \(\frac{4(x-2)}{x-y-3}=\log (x-2)+c\)
Reducible To Homogeneous Differential Equations Examples Explained
Example 4: Solve \((4 x+6 y+5) \frac{d y}{d x}=3 y+2 x+4\)
Solution:
Given Equation is \(\frac{d y}{d x}=\frac{2 x+3 y+4}{4 x+6 y+5}\) ………………………(1)
Here \(a_1=2, b_1=3, a_2=4, b_2=6 \Rightarrow a_1 b_2-a_2 b_1=12-12=0\)
Also \(a_2=2 a_1, b_2=2 b_1 \text { and } c_2 \neq 2 c_1\)
Hence Put \(2 x+3 y=u \Rightarrow 2+3 \frac{d y}{d x}=\frac{d u}{d x} \Rightarrow \frac{d y}{d x}=\frac{1}{3}\left(\frac{d u}{d x}-2\right)\) ………………………(2)
(1) and (2) => \(\frac{1}{3}\left(\frac{d u}{d x}-2\right)=\frac{u+4}{2 u+5} \Rightarrow \frac{d u}{d x}=\frac{3 u+12}{2 u+5}+2 \Rightarrow \frac{d u}{d x}=\frac{7 u+22}{2 u+5}\) ………………………..(3)
Separating the variables : \(\frac{2 u+5}{7 u+22} d u=d x \Rightarrow \int \frac{2 u+5}{7 u+22} d u=\int d x+c\)
⇒ \(\int\left(\frac{2}{7}-\frac{9}{7} \cdot \frac{1}{7 u+22}\right) d u=x+c \Rightarrow \frac{2}{7} u-\frac{9}{7} \cdot \frac{1}{7} \log (7 u+22)=x+c\)
∴ General solution of (3) is \(14 u-9 \log (7 u+22)=49 x+49 c\)
Hence the G. S. of (1) is \(14(2 x+3 y)-9 \log |(14 x+21 y+22)|=49 x+c_1\) [from (2)]
Common Mistakes In Solving Reducible To Homogeneous Differential Equations
Example.5 : Solve \((2 x+2 y+1) \frac{d y}{d x}=x+y+3\)
Solution.
Given equation is \(\frac{d y}{d x}=\frac{x+y+3}{2 x+2 y+1}\) …………………………..(1)
where \(a_1=1, b_1=1, c_1=3, a_2=2, b_2=2, c_2=1 \text { and } a_1 b_2-a_2 b_1=2-2=0\)
Also \(a_2=2 a_1, b_2=2 b_1 \text { and } c_2 \neq 2 c_1\)
Let \(x+y=u \Rightarrow 1+\frac{d y}{d x}=\frac{d u}{d x} \Rightarrow \frac{d y}{d x}=\frac{d u}{d x}-1\) ………………………….(2)
(1) and (2) \(\Rightarrow \frac{d u}{d x}-1=\frac{u+3}{2 u+1} \Rightarrow \frac{d u}{d x}=\frac{u+3}{2 u+1}+1=\frac{3 u+4}{2 u+1}\) ……………………………(3)
Separating the variables : \(\frac{2 u+1}{3 u+4} d u=d x\)
⇒ \(\int\left[\frac{2}{3}-\frac{5}{3(3 u+4)}\right] d u=\int d x+c \Rightarrow \frac{2 u}{3}-\frac{5}{9} \log (3 u+4)=x+c\)
G.S. of (3) is \(6 u-5 \log (3 u+4)=9 x+9 c\)
∴ The general solution of (1) is \(6(x+y)-5 \log (3 x+3 y+4)=9 x+c_1\)