Question
State True or False for the following:
Solution of $\frac{\text{xdy}}{\text{dx}}=\text{y}+\text{x}\tan\Big(\frac{\text{y}}{\text{x}}\Big)$ is $\sin\Big(\frac{\text{y}}{\text{x}}\Big)=\text{cx}.$

Answer

True.Solution:
We have
$\frac{\text{xdy}}{\text{dx}}=\text{y}+\text{x}\tan\Big(\frac{\text{y}}{\text{x}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}+\text{x}\tan\Big(\frac{\text{y}}{\text{x}}\Big)\ .....(\text{i})$
Put $\frac{\text{y}}{\text{x}}=\text{v}$ or $\text{y}=\text{vx}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{v}+\frac{\text{dv}}{\text{dx}}$
On substituting these values in Eq. (i), we get
$\frac{\text{xdy}}{\text{dx}}+\text{v}=\text{v}+\tan\text{v}$
$\Rightarrow\frac{\text{dx}}{\text{x}}=\frac{\text{dv}}{\tan\text{v}}$
On integrating both sides, we get
$\int\frac{1}{\text{x}}\text{dx}=\int\cot\text{v}\text{dv}$
$\Rightarrow\log\text{x}+\log\text{C}=\log\sin\text{v}$
$\Rightarrow\log\text{Cx}=\log\sin\text{v}$
$\Rightarrow\sin\text{v}=\text{Cx}$
$\Rightarrow\sin\frac{\text{y}}{\text{x}}=\text{C}\text{x}^4$

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