Q.$x=\cos t$, $y=e^{mt}$, show that $(1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}-m^2y=0$

Given $x=\cos t$, $y=e^{mt}$.

Step 1: $\dfrac{dx}{dt}=-\sin t$, $\dfrac{dy}{dt}=me^{mt}$, so $y_1=\dfrac{dy}{dx}=\dfrac{me^{mt}}{-\sin t}=-\dfrac{me^{mt}}{\sin t}$.

Step 2 — differentiate $y_1$ w.r.t. $t$ using the quotient rule: $\dfrac{d}{dt}(y_1)=-m\cdot\dfrac{me^{mt}\sin t-e^{mt}\cos t}{\sin^2t}=-\dfrac{me^{mt}(m\sin t-\cos t)}{\sin^2t}$.

Step 3 — divide by $dx/dt=-\sin t$ to get $y_2$: $y_2=\dfrac{me^{mt}(m\sin t-\cos t)}{\sin^3t}$.

Step 4 — substitute into the target, using $x=\cos t$ so $1-x^2=\sin^2t$:

$(1-x^2)y_2=\sin^2t\cdot\dfrac{me^{mt}(m\sin t-\cos t)}{\sin^3t}=\dfrac{me^{mt}(m\sin t-\cos t)}{\sin t}$ …

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Sign slip differentiating $-me^{mt}/\sin t$ with the quotient rule — both the numerator's and denominator's d …