Solve dy dx = (x+y)+ (x+y). [Hint: Substitute x + y = z] - Vidyadip

Question

Solve $\frac{\text{dy}}{\text{dx}}=\cos(\text{x}+\text{y})+\sin(\text{x}+\text{y}).$
[Hint: Substitute x + y = z]

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Answer

Given, $\frac{\text{dy}}{\text{dx}}=\cos(\text{x}+\text{y})+\sin(\text{x}+\text{y})\ .......(\text{i})$
Put $\text{x}+\text{y}=\text{z}$
$\Rightarrow1+\frac{\text{dy}}{\text{dx}}=\frac{\text{dz}}{\text{dx}}$
On substituting these values in Eq. (i), we get
$\Rightarrow\Big(\frac{\text{dz}}{\text{dx}}-1\Big)=\cos\text{z}+\sin\text{z}$
$\Rightarrow\frac{\text{dz}}{\text{dx}}=(\cos\text{z}+\sin\text{z}+1)$
$\Rightarrow\frac{\text{dz}}{\cos\text{z}+\sin\text{z} +1}=\text{dx}$
On integrating both sides, we get
$\Rightarrow\int\frac{\text{dz}}{\cos\text{z}+\sin\text{z} +1}=\int1\text{dx}$
$\Rightarrow\int\frac{\text{dz}}{\frac{1-\tan^2\frac{\text{z}}{2}+2\tan\frac{\text{z}}{2}+1+\tan^2\frac{\text{z}}{2}}{\Big(1+\tan^2\frac{\text{z}}{2}\Big)}}=\int\text{dx}$
$\Rightarrow\int\frac{\Big(1+\tan^2\frac{\text{z}}{2}\Big)\text{dz}}{2+2\tan^2\frac{\text{z}}{2}}=\int\text{dx}$
$\Rightarrow\int\frac{\sec^2\frac{\text{z}}{2}\text{dz}}{2\Big(1+\tan\frac{\text{z}}{2}\Big)}=\int\text{dx}$
Put $1+\tan\frac{\text{z}}{2}=\text{t}$
$\Rightarrow\Big(\frac{1}{2}\sec^2\frac{\text{z}}{2}\Big)\text{dz}=\text{dt}$
$\Rightarrow\int\frac{\text{dt}}{\text{t}}=\int\text{dx}$
$\Rightarrow\log|\text{t}|=\text{x}+\text{C}$
$\Rightarrow\log|1+\tan\frac{\text{z}}{2}|=\text{x}+\text{C}$
$\Rightarrow\log\begin{vmatrix}1+\tan\frac{(\text{x}+\text{y})}{2} \end{vmatrix}=\text{x}+\text{ C}$

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