Question
If $\text{y}=\text{x}+\tan\text{x},$ show that $\cos^2\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\text{y}+2\text{x}=0$
✓
Answer
$\text{y}=\text{x}+\tan\text{x},$
differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{dy}}{\text{dx}}=1+\sec^2\text{x}$
differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=0+2\sec^2\times\tan\text{x}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\sin\text{x}}{\cos^3\text{x}}$
$\Rightarrow\cos^2\times\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\tan\text{x}+2\text{x}-2\text{x}$
$\Rightarrow\cos^2\times\frac{\text{d}^2\text{y}}{\text{dx}^2}=2(\text{x}+\tan\text{x})-2\text{x}$
$\Rightarrow\cos^2\times\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{y}-2\text{x}$
$\Rightarrow\cos^2\times\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\text{y}+2\text{x}=0$
differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{dy}}{\text{dx}}=1+\sec^2\text{x}$
differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=0+2\sec^2\times\tan\text{x}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\sin\text{x}}{\cos^3\text{x}}$
$\Rightarrow\cos^2\times\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\tan\text{x}+2\text{x}-2\text{x}$
$\Rightarrow\cos^2\times\frac{\text{d}^2\text{y}}{\text{dx}^2}=2(\text{x}+\tan\text{x})-2\text{x}$
$\Rightarrow\cos^2\times\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{y}-2\text{x}$
$\Rightarrow\cos^2\times\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\text{y}+2\text{x}=0$
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