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Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Solve the following differential equations $\frac{\text{dy}}{\text{dx}}=\frac{2\text{x}(\log\text{x}+1)}{\sin\text{y+y}\cos\text{y}},$ given that $\text{y}=0,$ when $\text{x}=1.$
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Answer

$\frac{\text{dy}}{\text{dx}}=\frac{2\text{x}(\log\text{x}+1)}{\sin\text{y+y}\cos\text{y}},\text{y}=0$ at $\text{x}=1$
$\int(\sin\text{y+y}\cos\text{y})\text{dy}=\int2\text{x}(\log\text{x}+1)\text{dx}$
$\Rightarrow\int\sin\text{y dy}+\int\text{y}\cos\text{y dy}=\int2\text{x}\log\text{x dx}+2\int\text{x dx}$
$\Rightarrow-\cos\text{y}+\big[\text{y}\times\int\cos\text{y dy}-\int(1\times\int\cos\text{y dy})\text{dy}\big]\\=2\Big[\log\text{x}\int\text{x dx}-\int\Big(\frac{1}{\text{x}}\int\text{x dx}\Big)\text{dx}\Big]+\text{x}^2+\text{C}$
$\Rightarrow-\cos\text{y + y}\sin\text{y}-\int\sin\text{y dy}=2\frac{\text{x}^2}{2}\log\text{x}-2\int\frac{\text{x}}{2}\text{dx}+\text{x}^2+\text{C}$
$\Rightarrow-\cos\text{y + y}\sin\text{y}+\cos\text{y}=\text{x}^2\log\text{x}-\frac{\text{x}^2}{2}+\text{x}^2+\text{C}$
$\text{y}\sin\text{y}=\text{x}^2\log\text{x}+\frac{\text{x}^2}{2}+\text{C}$
Put $\text{y}=0,\text{x}=1$
$0=0+\frac{1}{2}+\text{C}$
$\text{C}=-\frac{1}{2}$
Put $\text{C}=-\frac{1}{2}$ in equation (1),
$\text{y}\sin\text{y = x}^2\log\text{x}+\frac{\text{x}^2}{2}-\frac{1}{2}$
$2\text{y}\sin\text{y}=2\text{x}^2\log\text{x + x}^2-1$

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