how can we solve ; dy/dx=tan 2(x+y) or dy/dx=tan square (x+y) - Maths - Differential Equations

Question

how can we solve ; dy/dx=tan 2(x+y) or dy/dx=tan square
(x+y)

Aakash Sharma · Accepted Answer

Dear Student!

Here is the answer to your query

You can solve $\frac{d y}{d x}$ = tan 2(x + y) or $\frac{d y}{d x}$ = tan²(x + y) by substitute x + y = z i.e.

Let x + y = z

$\Rightarrow 1 + \frac{d y}{d x} = \frac{d z}{d x} \Rightarrow \frac{d y}{d x} = \frac{d z}{d x} - 1$

Now $\frac{d y}{d x}$ = tan2 (x + y)

$\Rightarrow \frac{d z}{d x} - 1$ = tan 2 z

$\Rightarrow \frac{d z}{d x} = 1 + \tan 2 z$

$\Rightarrow \frac{d z}{1 + \tan 2 z} = d x$

$\Rightarrow \frac{1}{2} \int \frac{2 \cos 2 z}{\cos 2 z + \sin 2 z} d z = \int d x \Rightarrow \frac{1}{2} \int \frac{(\cos 2 z + \sin 2 z + \cos 2 z - \sin 2 z)}{\cos 2 z + \sin 2 z} d z = \int d x \Rightarrow \frac{1}{2} \int \frac{\cos 2 z + \sin 2 z}{\cos 2 z + \sin 2 z} d z + \frac{1}{2} \int \frac{\cos 2 z - \sin 2 z}{\cos 2 z + \sin 2 z} d z = \int d x \Rightarrow \frac{1}{2} \int d z + \frac{1}{2} \int \frac{\cos 2 z - \sin 2 z}{\cos 2 z + \sin 2 z} d z = \int d x$

Let cos 2 z + sin 2 z = v

$\Rightarrow - 2 \sin 2 z + 2 \cos 2 z = \frac{d v}{d z}$

$\Rightarrow d z (\cos 2 z - \sin 2 z) = \frac{d v}{2}$

$\Rightarrow \frac{1}{2} \int d z + \frac{1}{4} \int \frac{1}{v} d v = \int d x \Rightarrow \frac{z}{2} + \frac{1}{4} \log | v | = x + C \Rightarrow \frac{z}{2} + \frac{1}{4} \log | \cos 2 z + \sin 2 z | = x + C \Rightarrow \frac{x + y}{2} + \frac{1}{4} \log | \cos 2 z + \sin 2 z | = x + C \Rightarrow \frac{1}{4} \log | \cos 2 (x + y) + \sin 2 (x + y) | = \frac{x - y}{2} + C \Rightarrow \log | \cos 2 (x + y) + \sin 2 (x + y) | = 2 (x - y) + 4 C$

Try solving the other part yourself and if you face any difficulty then do get back to us.

Cheers!