Question

find the differential equation whose general solution is : y^n =ax^n​

Answer 1

Answer is given in the pic.

Answer 2

GIVEN :–

• Equation => yⁿ = axⁿ

TO FIND :–

• Differential equation = ?

SOLUTION :–

$\\ \bf \implies \: {y}^{n} = a {x}^{n}$

• Now Differentiate with respect to 'x' –

$\\ \bf \implies n({y}^{n - 1}) \dfrac{dy}{dx}=a(n){x}^{n - 1}$

$\\ \bf \implies{y}^{n - 1}\dfrac{dy}{dx}=a{x}^{n - 1}$

$\\ \bf \implies\dfrac{dy}{dx}=a \bigg( \dfrac{{x}^{n - 1}}{{y}^{n - 1}} \bigg)$

• Using given equation –

$\\ \bf \implies \: {y}^{n} = a {x}^{n}$

$\\ \bf \implies a = \dfrac{{y}^{n}}{{x}^{n}}$

$\\ \bf \implies\dfrac{dy}{dx}=\bigg( \dfrac{{y}^{n}}{{x}^{n}} \bigg) \bigg( \dfrac{{x}^{n - 1}}{{y}^{n - 1}} \bigg)$

$\\ \bf \implies\dfrac{dy}{dx}=\bigg( \dfrac{{y}^{ \{n - (n - 1) \}}}{{x}^{ \{n - (n - 1) \}}} \bigg)$

$\\ \bf \implies\dfrac{dy}{dx}=\bigg( \dfrac{{y}^{(n - n+1)}}{{x}^{(n - n + 1)}} \bigg)$

$\\ \bf \implies\dfrac{dy}{dx}=\bigg( \dfrac{{y}^{(1)}}{{x}^{(1)}} \bigg)$

$\\ \large \implies{ \boxed{ \bf\dfrac{dy}{dx}=\dfrac{y}{x}}}$