# Question 1b4c7

##### 1 Answer
Mar 17, 2017

$\left(1\right) :$The GS is, ${\left({y}^{2} - 8\right)}^{2} = k {e}^{a r c \tan \left(\frac{x}{4}\right)} .$

$\left(2\right) :$ The PS is, ${\left({y}^{2} - 8\right)}^{2} = {e}^{a r c \tan \left(\frac{x}{4}\right) - \frac{\pi}{4}} .$

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{y}^{2} - 8}{y {x}^{2} + 16 y} = \frac{{y}^{2} - 8}{y \left({x}^{2} + 16\right)} = \left\{\frac{{y}^{2} - 8}{y}\right\} \left\{\frac{1}{{x}^{2} + 16}\right\} .$

$\Rightarrow \frac{y}{{y}^{2} - 8} \mathrm{dy} = \frac{\mathrm{dx}}{{x}^{2} + 16} .$

This shows that the given Diff. Eqn. is of the Separable

Variable Type.

To find its General Solution (GS), we integrate it termwise.

$\therefore \int \frac{y}{{y}^{2} - 8} \mathrm{dy} + \ln c = \int \frac{1}{{x}^{2} + 16} \mathrm{dx} .$

;. 1/2int{d/dy(y^2-8)}/(y^2-8)+lnc=int1/{x^2+4^2)dx.#

$\therefore \frac{1}{2} \ln | {y}^{2} - 8 | + \ln c = \frac{1}{4} a r c \tan \left(\frac{x}{4}\right) .$

$\therefore 2 \ln | {y}^{2} - 8 | + 4 \ln c = a r c r a n \left(\frac{x}{4}\right) , i . e . ,$

$\ln {\left({y}^{2} - 8\right)}^{2} + \ln {c}^{4} = a r c \tan \left(\frac{x}{4}\right) , \mathmr{and} ,$

$\ln \left\{{c}^{4} {\left({y}^{2} - 8\right)}^{2}\right\} = a r c \tan \left(\frac{x}{4}\right)$

$\therefore {c}^{4} {\left({y}^{2} - 8\right)}^{2} = {e}^{a r c \tan \left(\frac{x}{4}\right)}$

Hence, the GS is, ${\left({y}^{2} - 8\right)}^{2} = k {e}^{a r c \tan \left(\frac{x}{4}\right)} , k = \frac{1}{c} ^ 4. . . \left(1\right)$

To find its Particular Solution (PS), we have to use the

Initial Condition $: y \left(x = 4\right) = 3 , \text{ i.e., when, } x = 4 , y = 3.$

Subst.ing in $\left(1\right) , {\left(9 - 8\right)}^{2} = k {e}^{a r c \tan \left(\frac{4}{4}\right)} , i . e . , 1 = k {e}^{\frac{\pi}{4}}$

$\therefore k = \frac{1}{e} ^ \left(\frac{\pi}{4}\right) = {e}^{- \frac{\pi}{4}} .$

Therefore, from $\left(1\right)$, we have the PS $: {\left({y}^{2} - 8\right)}^{2} = {e}^{a r c \tan \left(\frac{x}{4}\right) - \frac{\pi}{4}} .$

Enjoy Maths.!