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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations4 Marks
Question
Solve the following differential equation:
$(\text{x}^2+\text{y}^2)\frac{\text{dy}}{\text{dx}}=8\text{x}^2-3\text{xy}+2\text{y}^2$
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Answer

We have,

$(\text{x}^2+\text{y}^2)\frac{\text{dy}}{\text{dx}}=8\text{x}^2-3\text{xy}+2\text{y}^2$

$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{8\text{x}^2-3\text{xy}+2\text{y}^2}{\text{x}^2+\text{y}^2}$

This is a homogeneous differential equation.

Putting x = vy and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get

$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{8\text{x}^2-3\text{vx}^2+2\text{v}^2\text{x}^2}{\text{x}^2+\text{v}^2\text{x}^2}$

$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{8-3\text{v}+2\text{v}^2}{1+\text{v}^2}-\text{v}$

$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{8-4\text{v}+2\text{v}^2-\text{v}^3}{1+\text{v}^2}$

$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{4(2-\text{v})+\text{v}^2(2-\text{v})}{1+\text{v}^2}$

$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{(4+\text{v}^2)(2-\text{v})}{1+\text{v}^2}$

$\Rightarrow\ \frac{1+\text{v}^2}{(4+\text{v}^2)(2-\text{v})}\text{dv}=\frac{1}{\text{x}}\text{dx}$

 

Integrating both sides, we get

$\int\frac{1+\text{v}^2}{(4+\text{v}^2)(2-\text{v})}\text{dv}=\int\frac{1}{\text{x}}\text{dx}\ \dots(1)$

Let us consider the left hand side of (1).

Using partial fraction,

Let $\frac{1+\text{v}^2}{(4+\text{v}^2)(2-\text{v})}=\frac{\text{Av + B}}{4+\text{v}^2}+\frac{\text{C}}{2-\text{v}}$

$\Rightarrow\ 1+\text{v}^2=\text{Av}(2-\text{v})+\text{B}(2-\text{v})+\text{C}(4+\text{v}^2)$

$\Rightarrow\ 1+\text{v}^2=2\text{Av}-\text{Av}^2+2\text{B}-\text{Bv}+4\text{C}+\text{Cv}^2$

Comparing the co-efficients of both sides, we get

2A - B = 0

-A + C = 1

& 2B + 4C = 1

Solving these three equations, we get

$\text{A}=\frac{-3}8,\ \text{B}=\frac{-3}4$ and $\text{C}=\frac{5}8$

$\therefore\ \frac{1+\text{v}^2}{(4+\text{v}^2)(2-\text{v})}=\frac{-\frac{3}{8}\text{v}-\frac{3}{4}}{4+\text{v}^2}+\frac{\frac{5}{8}}{2-\text{v}}\ \dots(2)$

From (1) and (2), we get

$\int\frac{-\frac{3}{8}\text{v}-\frac{3}{4}}{4+\text{v}^2}+\frac{\frac{5}{8}}{2-\text{v}}=\int\frac{1}{\text{x}}\text{dx}$

$\Rightarrow\ -\frac{3}8\int\frac{\text{v}}{\text{v}^2+4}\text{dv}-\frac{3}4\int\frac{1}{\text{v}^2+4}\text{dv}+\frac{5}8\int\frac{1}{2-\text{v}}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$

$\Rightarrow\ \frac{-3}{16}\log|\text{v}^2+4|-\frac{3}{4\times2}\tan^{-1}\frac{\text{v}}2-\frac{5}8\log|2-\text{v}|=\log|\text{x}|+\log\text{C}$

$\Rightarrow\ -\frac{3}{4\times2}\tan^{-1}\frac{\text{v}}2=\log\bigg|\text{Cx}(2-\text{v})^{\frac{5}8}(\text{v}^2+4)^{\frac{3}{16}}\bigg|$

$\Rightarrow\ \text{e}^{-\frac{3}8\tan^{-1}\frac{\text{v}}2}=\text{C}\bigg|\text{x}(2-\text{v})^{\frac{5}8}(\text{v}^2+4)^{\frac{3}{16}}\bigg|$

Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get

$\Rightarrow\ \text{e}^{-\frac{3}8\tan^{-1}\frac{\text{y}}{2\text{x}}}=\text{C}\Bigg|\text{x}\Big(2-\frac{\text{y}}{\text{x}}\Big)^{\frac{5}8}\Big(\frac{\text{y}^2}{\text{x}^2}+4\Big)^{\frac{3}{16}}\Bigg|$

$\Rightarrow\ \text{e}^{-\frac{3}8\tan^{-1}\frac{\text{y}}{2\text{x}}}=\text{C}\bigg|\text{x}\times\frac{1}{\text{x}}(2\text{x}-\text{y})^{\frac{5}8}(\text{y}^2+4\text{x}^2)^{\frac{3}{16}}\bigg|$

$\Rightarrow\ \text{e}^{-\frac{3}8\tan^{-1}\frac{\text{y}}{2\text{x}}}=\text{C}|2\text{x}-\text{y}|^{\frac{5}8}(\text{y}^2+4\text{x}^2)^{\frac{3}{16}}$

Hence, $\text{e}^{-\frac{3}8\tan^{-1}\frac{\text{y}}{2\text{x}}}=\text{C}|2\text{x}-\text{y}|^{\frac{5}8}(\text{y}^2+4\text{x}^2)^{\frac{3}{16}}$ is the required solution.

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