# If the solve the problem

Question:

If $x=a \sin t-b \cos t, y=a \cos t+b \sin t$, prove that $\frac{d^{2} y}{d x^{2}}=-\frac{x^{2}+y^{2}}{y^{3}}$

Solution:

Formula: -

(i) $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}_{1}$ and $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{y}_{2}$

(ii) $\frac{\mathrm{d}}{\mathrm{dx}} \cos \mathrm{x}=\sin \mathrm{x}$

(iii) $\frac{d}{d x} \sin x=-\cos x$

(iv) $\frac{d}{d x} x^{n}=n x^{n-1}$

(v) chain rule $\frac{\mathrm{df}}{\mathrm{dx}}=\frac{\mathrm{d}(\mathrm{wou})}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\mathrm{dw}}{\mathrm{ds}} \cdot \frac{\mathrm{ds}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}$

(vi) parameteric forms $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

Given: -

$x=a \sin t-b c o s t, y=a c c o s t+b \sin t$

differentiating both w.r.t t

$\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{acost}+\mathrm{bsint} \frac{\mathrm{dy}}{\mathrm{dt}}=-\mathrm{asint}+\mathrm{bcost}$

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{y}, \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{x}$

Dividing both

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}} \times \frac{\mathrm{dt}}{\mathrm{dx}}=-\frac{\mathrm{x}}{\mathrm{y}}$

Differentiating w.r.t t

$\Rightarrow \frac{d\left(\frac{d y}{d x}\right)}{d t}=-\frac{y\left(\frac{d x}{d t}\right)-x\left(\frac{d y}{d t}\right)}{y^{2}}$

Putting the value

$\Rightarrow \frac{d\left(\frac{d y}{d x}\right)}{d t}=-\frac{\left\{y^{2}+x^{2}\right\}}{y^{2}}$

Dividing them

$\Rightarrow \frac{d^{2} y}{d x^{2}}=-\frac{\left\{y^{2}+x^{2}\right\}}{y^{2} \cdot y}=-\frac{\left\{x^{2}+y^{2}\right\}}{y^{3}}$

Hence proved.