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Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{2\text{y}+\text{x}}$
✓

Answer

We have,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{2\text{y}+\text{x}}$
This is a homogeneous differential equation.
Putting y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}}{2\text{vx + x}}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{1}{2\text{v}+1}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1}{2\text{v}+1}-\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1-2\text{v}^2-\text{v}}{2\text{v}+1}$
$\Rightarrow\ \frac{2\text{v}+1}{1-2\text{v}^2-\text{v}}\text{dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{2\text{v}+1}{1-2\text{v}^2-\text{v}}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{2\text{v}+1}{2\text{v}^2+\text{v}-1}\text{dv}=-\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{2\text{v}+1}{2\text{v}(\text{v}+1)-1(\text{v}+1)}\text{dv}=-\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{2\text{v}+1}{(2\text{v}-1)(\text{v}+1)}\text{dv}=-\int\frac{1}{\text{x}}\text{dx}\ \dots(1)$
Solving left hand side integral of (1), we get
Using partial fraction,
Let $\frac{2\text{v}+1}{(2\text{v}-1)(\text{v}+1)}=\frac{\text{A}}{(2\text{v}-1)}+\frac{\text{B}}{(\text{v}+1)}$
$\therefore\ \text{A}+2\text{B}=2\ \dots(2)$
And $\text{A}-\text{B}=1\ \dots(3)$
Solving (2) and (3), we get
$\text{A}=\frac{4}3$ and $\text{B}=\frac{1}3$
$\therefore\ \int\frac{2\text{v}+1}{(2\text{v}-1)(\text{v}+1)}\text{dy}=\frac{4}3\int\frac{1}{2\text{v}-1}\text{dv}+\frac{1}3\int\frac{1}{\text{v}+1}\text{dv}$
$=\frac{4}{3\times2}\log|2\text{v}-1|+\frac{1}3\log|\text{v}+1|+\log\text{C}$
From (1), we get
$\frac{2}3\log|2\text{v}-1|+\frac{1}3|\text{v}+1|+\log\text{C}=-\log|\text{x}|+\log\text{C}_1$
$\Rightarrow\ \log\Big\{\big|(2\text{v}-1)^2\big||\text{v}+1|\Big\}=-3\log|\text{x}|+\log\text{C}_2$
$\Rightarrow\ \log\Big\{\big|(2\text{v}-1)^2\big||\text{v}+1|\Big\}=\log\Big|\frac{\text{C}_2^3}{\text{x}^3}\Big|$
$\Rightarrow\ (2\text{v}-1)^2(\text{v}+1)=\frac{\text{C}_2^3}{\text{x}^3}$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\Rightarrow\ \Big(\frac{2\text{y}-\text{x}}{\text{x}}\Big)^2\Big(\frac{\text{y}+\text{x}}{\text{x}}\Big)=\frac{\text{C}_2^3}{\text{x}^3}$
$\Rightarrow\ (\text{x + y})(2\text{y}-\text{x})^2=\text{K}$

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