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Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Solve the following differential equation:
$(2\text{x}^2\text{y}+\text{y}^3)\text{dx}+(\text{xy}^2-3\text{x}^2)\text{dy}=0$
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Answer

$(2\text{x}^2\text{y}+\text{y}^3)\text{dx}+(\text{xy}^2-3\text{x}^2)\text{dy}=0$
$\frac{\text{dy}}{\text{dx}}=\frac{2\text{x}^2\text{y}+\text{y}^3}{\text{xy}^2-3\text{x}^2}$
It is a homogeneous equation
Put y = vx
and $\frac{\text{dy}}{\text{dx}}=\text{v +x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v +x}\frac{\text{dv}}{\text{dx}}=\frac{2\text{x}^2\text{vx}+\text{v}^3\text{x}^3}{3\text{x}^3-\text{xv}^2\text{x}^2}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{2\text{v + v}^3}{3-\text{v}^2}-\text{v}$
$=\frac{2\text{v + v}^3-3\text{v + v}^3}{3-\text{v}^2}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{2\text{v}^3-\text{v}}{3-\text{v}^2}$
$\int\frac{3-\text{v}^2}{2\text{v}^3-\text{v}}\text{dv}=\int\frac{\text{dx}}{\text{x}}\ \dots(\text{i})$
$\frac{3-\text{v}^2}{\text{v}(2\text{v}^2-1)}=\frac{\text{A}}{(\text{v})}+\frac{\text{Bv + C}}{(2\text{v}^2-1)}$
$3-\text{v}^2=\text{A}(2\text{v}^2-1)+(\text{Bv + C})(\text{v})$
$=2\text{Av}^2-\text{A}+\text{Bv}^2+\text{Cv}$
$3-\text{v}^2=(2\text{A + B})\text{v}^2\text{Cv}-\text{A}$
Comparing the co-efficient of like powers of v
A = -3
C = 0
and 2A + B = -1
⇒ 2(-3) + B = -1
⇒ B = 5
So,
$\int\frac{-3}{\text{v}}\text{dv}+\int\frac{5\text{v}}{2\text{v}^2-1}\text{dv}=\int\frac{\text{dx}}{\text{x}}$
$-3\int\frac{1}{\text{v}}\text{dv}+\frac{5}4\int\frac{4\text{v}}{2\text{v}^2-1}\text{dv}=\int\frac{\text{dx}}{\text{x}}$
$-3\log|\text{v}|+\frac{5}4\log|2\text{v}^2-1|=\log|\text{x}|+\log|\text{C}|$
$-12\log|\text{v}|+5\log|2\text{v}^2-1|=4\log|\text{x}|+4\log|\text{C}$
$\frac{|2\text{v}^2-1|^5}{\text{v}^{12}}=\text{x}^4\text{C}^4$
$\frac{|2\text{y}^2-\text{x}^2|^5}{\text{x}^{10}}=\text{x}^4\text{C}^4\Big(\frac{\text{y}}{\text{x}}\Big)^{12}$
$|2\text{y}^2-\text{x}^2|^5=\text{x}^{14}\text{C}^4\frac{\text{y}^{12}}{\text{x}^{12}}$
$\text{x}^2\text{C}^4\text{y}^{12}=|2\text{y}^2-\text{x}^2|^5$

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