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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations4 Marks
Question
Solve the following initial value problems:
$(\text{y}^4-2\text{x}^3\text{y})\text{dx}+(\text{x}^4-2\text{xy}^3)\text{dy}=0,\text{y}(1)=1$
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Answer

$(\text{y}^4-2\text{x}^3\text{y})\text{dx}+(\text{x}^4-2\text{xy}^3)\text{dy}=0,\text{y}(1)=1$
This is a homogeneous equation, put y = vx
$\big(\text{v}^4\text{x}^4-2\text{vx}^4\big)+\big(\text{x}^4-2\text{v}^3\text{x}^4\big)\Big[\text{v + x}\frac{\text{dv}}{\text{dx}}\Big]=0$
$\big(\text{v}^4\text{x}^4-2\text{vx}^4\big)=\big(2\text{v}^3\text{x}^4-\text{x}^4\big)\Big[\text{v + x}\frac{\text{dv}}{\text{dx}}\Big]$
$\text{vx}^4(\text{v}^3-2)=\text{x}^4(2\text{v}^3-1)\Big[\text{v + x}\frac{\text{dv}}{\text{dx}}\Big]$
$\text{v}(\text{v}^3-2)=(2\text{v}^3-1)\text{v + x}(2\text{v}^3-1)\frac{\text{dv}}{\text{dx}}$
$\text{v}\big[\text{v}^3-2-2\text{v}^3+1\big]=\text{x}(2\text{v}^3-1)\frac{\text{dv}}{\text{dx}}$
$\text{v}(-1-\text{v}^3)=\text{x}(2\text{v}^3-1)\frac{\text{dv}}{\text{dx}}$
$\text{v}(1+\text{v}^3)=\text{x}(1-2\text{v}^3)\frac{\text{dv}}{\text{dx}}$
$\frac{\text{dx}}{\text{x}}=\frac{(1-2\text{v}^3)}{\text{v}(1+\text{v}^3)}\text{dv}$
On integrating both side of the equation we get,
$\int\frac{\text{dx}}{\text{x}}=\int\frac{(1-2\text{v}^3)}{\text{v}(1+\text{v}^3)}\text{dv}$
$\Rightarrow\ \log_{\text{e}}\text{x}=\int\frac{1+\text{v}^3-3\text{v}^3}{\text{v}(1+\text{v}^3)}\text{dv}$
$\Rightarrow\ \log_{\text{e}}\text{x}=\int\frac{1+\text{v}^3}{\text{v}(1+\text{v}^3)}\text{dv}-\int\frac{3\text{v}}{\text{v}(1+\text{v}^3)}\text{dv}$
$\Rightarrow\ \log_{\text{e}}\text{x}=\int\frac{1}{\text{v}}\text{dv}-\int\frac{3\text{v}^2}{(1+\text{v}^3)}\text{dv}$
$\Rightarrow\ \log_{\text{e}}\text{x}=\log_{\text{e}}\text{v}-\int\frac{\text{dt}}{\text{t}}$
$\Rightarrow\ \log_{\text{e}}\text{x}=\log_{\text{e}}\text{v}-\log_{\text{e}}(1+\text{v}^3)+\text{C}$ let $(1+\text{v}^3)=\text{t},3\text{v}^2\text{dv}=\text{dt}$
$\Rightarrow\ \log_{\text{e}}\text{x}=\log_{\text{e}}\frac{\text{v}}{1+\text{v}^3}+\text{C}$
As $\text{v}=\frac{\text{y}}{\text{x}}$
$\Rightarrow\ \log_{\text{e}}\text{x}=\log_{\text{e}}\frac{\frac{\text{y}}{\text{x}}}{1+\text{y}^{\frac{3}{\text{x}}}}+\text{C}$
$\Rightarrow\ \log_{\text{e}}\text{x}=\log_{\text{e}}\frac{\text{yx}^2}{\text{x}^3+\text{y}^3}+\text{C}$
As y(1) = 1
$\Rightarrow\ \log_{\text{e}}1=\log_{\text{e}}\frac{1}{1+1}+\text{C}$
$\Rightarrow\ 0=\log_{\text{e}}\frac{1}2+\text{C}$
$\text{C}=-\log_{\text{e}}\frac{1}2$
$\Rightarrow\ \text{C}=\log_{\text{e}}2$
$\therefore\ \log_{\text{e}}\text{x}=\log_{\text{e}}\frac{\text{yx}^2}{\text{x}^3+\text{y}^3}+\log_{\text{e}}2$

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