Question

Solve the differential equation

$x \frac{dy}{dx} = y - xtan( \frac{y}{x}) \: given \: that \: y(1) = \frac{\pi}{4}$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given differential equation is

$\rm \: x\dfrac{dy}{dx} = y - xtan\bigg(\dfrac{y}{x} \bigg)$

can be rewritten as

$\rm \: \dfrac{dy}{dx} = \frac{y}{x} - tan\bigg(\dfrac{y}{x} \bigg) - - - (1)$

Now, to evaluate this Differential equation, we use method of Substitution

So, Substitute y = vx ----(2)

On substituting the value in equation (1), we get

$\rm \: \dfrac{d}{dx}(vx) = \frac{vx}{x} - tan\bigg(\dfrac{vx}{x} \bigg)$

$\rm \: v\dfrac{d}{dx}x + x\dfrac{d}{dx}v = v - tanv$

$\rm \: v \times 1 + x\dfrac{dv}{dx} = v - tanv$

$\rm \: v + x\dfrac{dv}{dx} = v - tanv$

$\rm \: x\dfrac{dv}{dx} = - tanv$

On separating the variables, we get

$\rm \: \dfrac{dv}{tanv} = - \frac{dx}{x}$

$\rm \: cotv \: dv = - \frac{dx}{x}$

On integrating both sides w. r. t. x, we get

$\rm \: \displaystyle\int\rm cotv \: dv = - \displaystyle\int\rm \frac{dx}{x}$

$\rm \: log(sinv) = - log(x) + log(c)$

$\rm \: log(sinv) + log(x) = log(c)$

$\rm \: log(x \: sinv) = log(c)$

$\:\rm \: x \: sinv = c$

$\:\rm \: x \: sin\bigg(\dfrac{y}{x} \bigg) = c - - - (3)$

Now, it is given that $\rm \: x=1\:and\:y=\dfrac{\pi}{4}$

So, on substituting these values in above equation (3), we get

$\rm \: 1 \times sin\bigg(\dfrac{\pi}{4} \bigg) = c$

$\bf\implies \:c \: = \: \dfrac{1}{ \sqrt{2} }$

So, on substituting the value of c, in equation (3),

The particular solution is given by

$\bf\implies \:\:\bf \: x \: sin\bigg(\dfrac{y}{x} \bigg) = \dfrac{1}{ \sqrt{2} }$

$\rule{190pt}{2pt}$

Formulae Used :-

$\boxed{ \rm{ \:\dfrac{d}{dx}(uv) = u\dfrac{d}{dx}v \: + \: v\dfrac{d}{dx}u \: }}$

$\boxed{ \rm{ \:\displaystyle\int\rm \frac{1}{x} \: dx \: = \: log(x) + c \: }}$

$\boxed{ \rm{ \:\displaystyle\int\rm cotx \: dx \: = \: log(sinx) + c \: }}$

$\boxed{ \rm{ \: log(x) + log(y) = log(xy) \: }}$