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

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

Answers (1)

$$ \frac{d y}{d x}=\cos (x+y)+\sin (x+y) $$
Using the given hint substitute x+y=z
$$ \Rightarrow \frac{\mathrm{d}(\mathrm{z}-\mathrm{x})}{\mathrm{dx}}=\cos \mathrm{z}+\sin \mathrm{z} $$
Differentiate z- x with respect to x
$$ \\ \Rightarrow \frac{d z}{d x}-1=\cos z+\sin z \\ \Rightarrow \frac{d z}{d x}=1+\cos z+\sin z \\ \Rightarrow \frac{d z}{1+\cos z+\sin z}=d x $$

Integrate
$$ \Rightarrow \int \frac{\mathrm{d} z}{1+\cos z+\sin z}=\int \mathrm{d} \mathrm{x} $$
As we know
\cos 2 z=2 \cos ^{2} z-1$
And \sin 2 z=2 \sin z \cos z$
$$ \\ \Rightarrow \int \frac{d z}{1+2 \cos ^{2} \frac{z}{2}-1+2 \sin \frac{z}{2} \cos \frac{z}{2}}=x \\ \Rightarrow \int \frac{d z}{2 \cos ^{2} \frac{z}{2}+2 \sin \frac{z}{2} \cos \frac{z}{2}}=x \\ \Rightarrow \int \frac{d z}{2 \cos \frac{z}{2}\left(\cos \frac{z}{2}+\sin \frac{z}{2}\right)}=x \\ \Rightarrow \int \frac{d z}{2 \cos ^{2} \frac{z}{2}\left(1+\frac{\sin \frac{z}{2}}{\cos \frac{2}{2}}\right)}=x $$

\int \frac{\sec^2 \frac{z}{2}}{2(1+\tan \frac{z}{2})}dz

1+\tan \frac{z}{2}=t$
Differentiate with respect to z
We get
$$ \frac{d t}{d z}=\frac{\sec ^{2} \frac{z}{2}}{2} $$
hence \frac{\sec ^{2} \frac{z}{2} \mathrm{dz}}{2}=\mathrm{dt}$
\Rightarrow \int \frac{d t}{t}=x$

logt +c=x$
Again substitute t
$$ \Rightarrow \log \left(1+\tan \frac{z}{2}\right)+c=x $$
Similarly substitute z
$$ \Rightarrow \log \left(1+\tan \frac{x+y}{2}\right)+c=x $$

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infoexpert22

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