JEE Main 2022 — Mathematics Calculus

Given,

$${\mathrm{cos}}^{-1}(\frac{y}{2})={\mathrm{log}}_{e}{(\frac{x}{5})}^{5}$$cos−1(2y​)=loge​(5x​)5

Using property $\mathrm{log}{x}^{m}=m\mathrm{log}x$logxm=mlogx we get,

$${\mathrm{cos}}^{-1}(\frac{y}{2})=5{\mathrm{log}}_{e}(\frac{x}{5})$$cos−1(2y​)=5loge​(5x​)

Now differentiating both side w.r.t $x$x we get,

$$\frac{-1}{\sqrt{1-\frac{{y}^{2}}{4}}}\cdot \frac{{y}^{'}}{2}=5\cdot \frac{1}{\frac{x}{5}}\cdot (\frac{1}{5})$$1−4y2​​−1​⋅2y′​=5⋅5x​1​⋅(51​)

$$\Rightarrow \frac{-{y}^{'}}{\sqrt{4-{y}^{2}}}=\frac{5}{x}$$⇒4−y2​−y′​=x5​

$$\Rightarrow -x{y}^{'}=5\sqrt{4-{y}^{2}}$$⇒−xy′=54−y2​

Again differentiating w.r.t $x$x we get,

$$\Rightarrow -x{y}^{''}-{y}^{'}=5\cdot \frac{1}{2\sqrt{4-{y}^{2}}}(-2y{y}^{'})$$⇒−xy′′−y′=5⋅24−y2​1​(−2yy′)

$$\Rightarrow x{y}^{''}+{y}^{'}=\frac{5yy'}{\sqrt{4-{y}^{2}}}$$⇒xy′′+y′=4−y2​5yy′​

$$\Rightarrow x{y}^{''}+{y}^{'}=(5y)\cdot (-\frac{5}{x})$$⇒xy′′+y′=(5y)⋅(−x5​)

$$\Rightarrow {x}^{2}{y}^{''}+x{y}^{'}=-25y$$⇒x2y′′+xy′=−25y

$$\Rightarrow {x}^{2}{y}^{''}+x{y}^{'}+25y=0$$⇒x2y′′+xy′+25y=0