Question

If y = x + tan x, show that ${\cos }^{2}x\frac{d^{2}y}{d{x}^2}-2y+2x=0$.

Answer 1

Here,

$$\begin{gathered} y=x+\tan x \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}y}{\mathit{d}x}=1+{\sec }^{2}x \\ \mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^2}=2{\sec }^{2}x\tan x \\ \mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}{\sec }^{2}x,\mathrm{we}\mathrm{get} \\ {\cos }^{2}x\frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^2}=2\tan x \\ \Rightarrow {\cos }^{2}x\frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^2}=2(y-x)\left[\because y=x+\tan x\Rightarrow \tan x=y-x\right] \\ \Rightarrow {\cos }^{2}x\frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^2}-2y+2x=0 \end{gathered}$$

Hence proved.