Name: General Solution Using Variable Separable Method
Uploaded: 2019-07-19
Duration: 7 min 52 s
Description: We have seen what differential equations are, and we saw a few examples of how to find the general and particular solution of the ODE. Let us learn more.

Question

Solve :

\frac{d y}{d x} = \cos (x + y) + \sin (x + y)

Accepted Answer

Given, $\frac{d y}{d x} = \cos (x + y) + \sin (x + y)$ ...(i)

Put $x + y = z$

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

On substituting these values in Eq. (i), we get

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

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

$\Rightarrow \frac{d z}{\cos z + \sin z + 1} = d x$

On integrating both sides, we get

$\Rightarrow \int \frac{d x}{\cos z + \sin z + 1} = \int 1 d x$

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

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

$\Rightarrow \int \frac{(1 + \tan^{2} \frac{z}{2}) d z}{2 + 2 \tan^{2} \frac{z}{2}} = \int d x$

$\Rightarrow \int \frac{\sec^{2} \frac{z}{2} d z}{2 (1 + \tan \frac{z}{2})} = \int d x$

Put $1 + \tan \frac{z}{2} = t \Rightarrow (\frac{1}{2} \sec^{2} \frac{z}{2}) d z = d t$

$\Rightarrow \int \frac{d t}{t} = \int d x$

$\Rightarrow \log | t | = x + C$

$\Rightarrow \log | 1 + \tan \frac{z}{2} | = x + C$

$∴ \log |1 + \tan \frac{(x + y)}{2}| = x + C$ is the required solution.

EXERCISE

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