The general solution of $\frac{d y}{d x}=\cos ^2(x-y-1)$ is given by $\mathrm{x}=$
The general solution of $\frac{d y}{d x}=\cos ^2(x-y-1)$ is given by $\mathrm{x}=$
- $\mathrm{C}-\cot (\mathrm{x}-\mathrm{y}-1)$
- $\mathrm{C}-\tan (\mathrm{x}-\mathrm{y}+1)$
- $\mathrm{y}+\mathrm{C} \cot (\mathrm{x}-\mathrm{y}-1)$
- $\mathrm{Cy}+\tan (\mathrm{x}-\mathrm{y}-1)$
Solution
$\because \frac{d y}{d x}=\cos ^2(x-y-1)$
let $x-y-1=P \Rightarrow 1-\frac{d y}{d x}=\frac{d p}{d x} \Rightarrow \frac{d y}{d x}=1-\frac{d p}{d x}$
Using above substitution in $\mathrm{eq}^{\mathrm{n}}$ (i), we get
$\begin{aligned}
& 1-\frac{d p}{d x}=\cos ^2 p \Rightarrow \frac{d p}{d x}=1-\cos ^2 p \Rightarrow \frac{d p}{\sin ^2 p}=d x \\
& \Rightarrow \operatorname{cosec}^2 p d p=d x
\end{aligned}$
Integrating both sides, we get :-
$\begin{aligned}
& \Rightarrow-\cot \mathrm{p}=\mathrm{x}+\mathrm{c}^{\prime} \Rightarrow \mathrm{x}=-\mathrm{c}^{\prime}-\cot \mathrm{p} \\
& \Rightarrow \mathrm{x}=\mathrm{c}-\cot \mathrm{p} \quad \text { where } \mathrm{c}^{\prime}=-\mathrm{c} \\
& \Rightarrow \mathrm{x}=\mathrm{c}-\cot (\mathrm{x}-\mathrm{y}-1)
\end{aligned}$
Asked in: MHT CET Full Test 7
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