Question

differentiate Homogeneous differential equations dy/dx=y/x+tany/x
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Answer 1

$\bf \huge \: Answer$

$\huge{ \sf{ \underline{ \pink{we \: have,}}}}$

$\bf \frac{dy}{dx} = \tan( \frac{y}{x} ) \: + \: \frac{y}{x} ......(1)$

This is clearly a Homogenous differential equation, as RHS is expressed only in terms of $\bf\frac{y}{x}$

$\huge{ \sf{ \underline{ \pink{To \: Solve \: this,}}}}$

$\bf \: Let \: \frac{y}{x} \: = \: t \:$

$\bf \: \implies \: y \: = \: xt \:$

$\bf \: \implies \: \frac{dy}{dx} = \: t \: + \: x \frac {dt}{dx}$

$\bf \: By (i)$

$\bf \implies \: tan \: t \: = x \frac{dt}{dx}$

$\bf \implies \: tan \: t \: = \: x \frac{dt}{dx}$

$\bf \implies \: \frac{dx}{x} = \frac{dt}{tan \: t \: }$

$\huge{ \sf{ \underline{ \pink{Integrating \: both \: sides}}}}$

$\bf \implies \int \: \frac{dx}{x} \: = \: \int \: cot \: t \: dt \:$

$\bf \implies \: log \: x \: = \: log \bigg(sin \: t \bigg) \: + \: log \: C$

$\bf \implies \: log \: x \: = \: log \: \bigg( \: C \: sin \: \frac{y}{x} \bigg)$

$\bf \implies \boxed { x\: = \: C \: sin \: \bigg( \: \frac{y}{x} \bigg)}$

This is the solution of given Differential equation, But it could be solved further to get y.

$\bf \implies \: \frac{x}{C} \: = \: sin \: \bigg( \frac{y}{x} \bigg)$

$\bf \implies \: sin^{ - 1} \bigg(Ax \bigg) \: \frac{y}{x} \: where \: A = \frac{1}{C}$

$\bf \implies \boxed{y \: = \: x \: sin^{ - 1} \bigg(Ax \bigg)}$

$\bf \: Hope \: it's \: helps :)$