Question

Verify that the given function is a solution of the differential equation, $y=x^{m}; x^{2}\frac{d^{2}y}{dx^{2}}-mx \frac{dy}{dx}+my=0$

Answer 1

To verify that the given function is a solution of the differential equation, $y=x^{m}; x^{2}\frac{d^{2}y}{dx^{2}}-mx \frac{dy}{dx}+my=0$

take second order derivative ,and manage terms to get the required differential equation

$y = {x}^{m}...eq1 \\ \\ \frac{dy}{dx} = m \: {x}^{m - 1} ...eq2 \\ \\ \frac{ {d}^{2}y }{ {dx}^{2} } = m(m - 1) {x}^{m - 2} ...eq3$
put all these equation in

$x^{2}\frac{d^{2}y}{dx^{2}}-mx \frac{dy}{dx}+my=0$

${x}^{2} m(m - 1) {x}^{m - 2} - mx(m \: {x}^{m - 1} ) + m \: {x}^{m} \\ \\ {x}^{m - 2+2}( {m}^{2} - m) - {m}^{2} x \: {x}^{m - 1} + m \: {x}^{m} \\ \\ {x}^{m} {m}^{2} - m {x}^{m} - {m}^{2} {x}^{m - 1 + 1} + m {x}^{m} \\ \\ {x}^{m} {m}^{2} - m {x}^{m } - {m}^{2} {x}^{m} + m {x}^{m} \\ \\ = 0$

hence proved