Question

In each of the Exercise 1 to 10, show that the given differential equation is homogeneous and solve :-

$\sf {(x^2\ +\ xy)dy\ =\ (x^2\ +\ y^2)dx}$

Answer 1

Step-by-step explanation:

Question :-

In each of the Exercise 1 to 10, show that the given differential equation is homogeneous and solve :-

$\sf {(x^2\ +\ xy)dy\ =\ (x^2\ +\ y^2)dx}$

Solution :-

$\rm \circ\ {Step\ 1\ :\ Find\ \dfrac{dy}{dx}}$

$: \implies {(x^2\ +\ xy)dy\ =\ (x^2\ +\ y^2)dx}$

${: \implies \dfrac{dy}{dx}\ =\ \dfrac{x^2\ +\ y^2}{x^2\ +\ xy}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: -\ -\ -\ -\ -\ -\ (1)}$

$\rm \circ\ {Step\ 2\ :\ Putting\ F(x,\ y)\ =\ \dfrac{dy}{dx}\ and\ finding\ F( \lambda x,\ \lambda y)}$

$: \implies {F(x,\ y)\ =\ \dfrac{x^2\ +\ y^2}{x^2\ +\ xy}}$

Finding F(λx, λy) :-

${: \implies F( \lambda x,\ \lambda y)\ =\ \dfrac{( \lambda x)^2\ +\ ( \lambda y)^2}{( \lambda x)^2\ +\ \lambda x\ +\ \lambda y}}$

$: \implies {\dfrac{ \lambda^2 x^2\ +\ \lambda^2 y^2}{ \lambda^2 x^2\ +\ \lambda^2 xy}}$

$: \implies {\dfrac{ \lambda^2 (x^2 y^2)}{ \lambda^2 (x^2\ +\ xy)}}$

$: \implies {\dfrac{x^2 y^2}{x^2\ +\ xy}}$

$: \implies {F(x,\ y)}$

${So,\ F( \lambda x,\ \lambda y)\ =\ F(x,\ y)}$

$: \implies { \lambda^0\ F(x,\ y)}$

Thus, F(x,y) is a homogeneous equation function of order zero.

$\mathrm {Therefore,\ \dfrac{dy}{dx}\ is\ a\ homogeneous\ differential\ equation.}$

$\rm \circ\ {Step\ 3\ :\ Solving\ \dfrac{dy}{dx}\ by\ putting\ y\ =\ vx}$

Differentiating w.r.t.x :-

$: \implies {\dfrac{dy}{dx}\ =\ x \dfrac{dv}{dx}\ +\ v \dfrac{dx}{dx}}$

$: \implies {\dfrac{dy}{dx}\ =\ x \dfrac{dv}{dx}\ +\ v}$

$\rm {Putting\ value\ of\ \dfrac{dy}{dx}\ and\ y\ =\ vx\ in\ (1)\ :-}$

$: \implies {\dfrac{dy}{dx}\ =\ \dfrac{x^2\ +\ y^2}{x^2\ +\ xy}}$

$: \implies {x \dfrac{dy}{dx}\ +\ v\ =\ \dfrac{x^2\ +\ (vx)^2}{x^2\ +\ x(vx)}}$

$: \implies {x \dfrac{dy}{dx}\ +\ v\ =\ \dfrac{x^2(1\ +\ v^2)}{x^2\ +\ x^2 v}}$

$: \implies {x \dfrac{dy}{dx}\ +\ v\ =\ \dfrac{x^2(1\ +\ v^2)}{x^2(1\ +\ v)}}$

$: \implies {x \dfrac{dy}{dx}\ +\ v\ =\ \dfrac{1\ +\ v^2}{1\ +\ v}}$

$: \implies {x \dfrac{dy}{dx}\ =\ \dfrac{1\ +\ v^2\ -\ v\ -\ v^2}{1\ +\ v}}$

$: \implies {x \dfrac{dy}{dx}\ =\ \dfrac{1\ -\ v}{1\ +\ v}}$

$: \implies {\dfrac{(1\ +\ v)}{(1\ -\ v)}\ dv\ =\ \dfrac{dx}{x}}$

$: \implies {- \bigg(\dfrac{v\ +\ 1}{v\ -\ 1} \bigg)\ dv\ =\ \dfrac{dx}{x}}$

$: \implies {\bigg(\dfrac{v\ +\ 1}{v\ -\ 1} \bigg)\ dv\ =\ \dfrac{-dx}{x}}$

By integrating both sides, we get :-

${: \implies \displaystyle \int\ \bigg(\dfrac{v\ +\ 1}{v\ -\ 1} \bigg)\ dv\ =\ -\ \displaystyle \int\ \dfrac{dx}{x}}$

${: \implies \displaystyle \int\ \bigg(\dfrac{v\ +\ 1}{v\ -\ 1} \bigg)\ dv\ =\ -log \left|x \right|\ +\ c\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: -\ -\ -\ -\ -\ -\ (2)}$

$: \implies {Let\ \bf I\ =\ \displaystyle \int\ \bigg(\dfrac{v\ +\ 1}{v\ -\ 1} \bigg)\ dv}$

Now, solving I :-

$: \implies {I\ =\ \displaystyle \int\ \bigg(\dfrac{v\ +\ 1\ -\ 1\ +\ 1}{v\ -\ 1} \bigg)\ dv}$

$: \implies {I\ =\ \displaystyle \int\ \bigg(\dfrac{v\ -\ 1\ +\ 2}{v\ -\ 1} \bigg)\ dv}$

$: \implies {I\ =\ \displaystyle \int\ \bigg(\dfrac{v\ -\ 1}{v\ -\ 1}\ +\ \dfrac{2}{v\ -\ 1} \bigg)\ dv}$

$: \implies {I\ =\ \displaystyle \int\ \bigg(1\ +\ \dfrac{2}{v\ -\ 1} \bigg)\ dv}$

$: \implies {I\ =\ \displaystyle \int\ dv\ +\ \displaystyle \int\ \dfrac{2}{v\ -\ 1}\ dv}$

$: \implies {I\ =\ \sf v\ +\ 2\ log \left|v\ -\ 1 \right|}$

Now, putting v = y/x :-

$: \implies {I\ =\ \dfrac{y}{x}\ +\ 2\ log \left| \dfrac{y}{x}\ -\ 1 \right|}$

$: \implies {I\ =\ \dfrac{y}{x}\ +\ 2\ log \left| \dfrac{y\ -\ x}{x} \right|}$

Now, putting value of I in (2), we get :-

${: \implies \dfrac{y}{x}\ +\ 2\ log \left| \dfrac{y\ -\ x}{x} \right|\ =\ -\ log \left|x \right|\ +\ c}$

${: \implies \dfrac{y}{x}\ +\ 2\ log \left| \dfrac{y\ -\ x}{x} \right|\ +\ log \left|x \right|\ =\ c}$

${: \implies \dfrac{y}{x}\ +\ 2\ log \left| \dfrac{(y\ -\ x)^2}{x^2} \right|\ +\ log \left|x \right|\ =\ c\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (As\ a\ log\ b\ =\ log\ b^a)}$

${: \implies \dfrac{y}{x}\ +\ 2\ log \left| \dfrac{(y\ -\ x)^2}{x}\ \times x \right|\ =\ c\: \: \: \: \: \: \: \: \: \: \: \: \: (As\ log\ a\ +\ log\ b\ =\ log\ ab)}$

${: \implies \dfrac{y}{x}\ +\ 2\ log \left| \dfrac{(y\ -\ x)^2}{x} \right|\ =\ c}$

${: \implies log \left| \dfrac{(y\ -\ x)^2}{x} \right|\ =\ c\ -\ \dfrac{y}{x}}$

${: \implies \dfrac{(y\ -\ x)^2}{x}\ =\ e^{c\ -\ \dfrac{y}{x}}}$

${: \implies \dfrac{(y\ -\ x)^2}{x}\ =\ e^c \times e^{- \dfrac{y}{x}}}$

${: \implies \dfrac{(y\ -\ x)^2}{x}\ =\ c\ e^{- \dfrac{y}{x}}}$

${: \implies (x\ -\ y)^2\ =\ cx\ e^{- \dfrac{y}{x}}}$

Answer 2

Answer:

$the \: above \: answer \: is \: correct$