Solve in two different ways (2)

**alexmahone** · September 26th 2011, 11:05 PM

Solve in two different ways.

$\frac{dy}{dx}=\frac{\sqrt{y}-y}{\tan x}$

Could someone please check the finer details of my solution, such as signs of integration constants and modulus signs?

Separable:

$\frac{dy}{\sqrt{y}-y}=\frac{dx}{\tan x}$

$\frac{1}{2\sqrt{y}}\frac{2}{1-\sqrt{y}}dy=\cot xdx$

$\ln |\sin x|=\int \frac{1}{2\sqrt{y}}\frac{2}{1-\sqrt{y}}dy$

Let $u=\sqrt{y}$ .

$du=\frac{dy}{2\sqrt{y}}$

$\int \frac{1}{2\sqrt{y}}\frac{2}{1-\sqrt{y}}dy=\int \frac{2}{1-u}du$

$=-2\ln |1-u|+C_1$

$=-2\ln |1-\sqrt{y}|+C_1$

$\ln |\sin x|=-2\ln |1-\sqrt{y}|+C_1$

$\ln |\sin x|(1-\sqrt{y})^2=C_1$

$|\sin x|(1-\sqrt{y})^2=e^{C_1}=C_2$ , where $C_2$ is a positive constant.

$(1-\sqrt{y})^2=C_2|\sin x|^{-1}$

$1-\sqrt{y}=\pm \sqrt{C_2}|\sin x|^{-1/2}$

$\sqrt{y}=1\mp \sqrt{C_2}|\sin x|^{-1/2}$

$\sqrt{y}=1+C|\sin x|^{-1/2}$ , where $C=\mp \sqrt{C_2}$ .

$y(x)=1+C^2|\sin x|^{-1}+2C|\sin x|^{-1/2}$

Bernoulli:

$\frac{dy}{dx}+\frac{y}{\tan x}=\frac{\sqrt{y}}{\tan x}$

Let $u=y^{1/2}$ .

$y=u^2$

$\frac{dy}{dx}=2u\frac{du}{dx}$

$2u\frac{du}{dx}+\frac{u^2}{\tan x}=\frac{u}{\tan x}$

$\frac{du}{dx}+\frac{u}{2\tan x}=\frac{1}{2\tan x}$

$\frac{du}{dx}+\frac{u\cot x}{2}=\frac{\cot x}{2}$

The integrating factor is $\exp\left(\int \frac{\cot x}{2}\right)=\exp\left(\frac{1}{2}\ln |\sin x|\right)=|\sin x|^{1/2}$ .

$|\sin x|^{1/2}\frac{du}{dx}+|\sin x|^{1/2}\frac{u\cot x}{2}=|\sin x|^{1/2}\frac{\cot x}{2}$

$\frac{d}{dx}(u|\sin x|^{1/2})=|\sin x|^{1/2}\frac{\cot x}{2}$

$u|\sin x|^{1/2}=|\sin x|^{1/2}+C$

$u=1+C|\sin x|^{-1/2}$

$y^{1/2}=1+C|\sin x|^{-1/2}$

$y(x)=1+C^2|\sin x|^{-1}+2C|\sin x|^{-1/2}$

**Ackbeet** · September 27th 2011, 07:34 AM

The beauty of DE's is it's often much easier to check your own solution than to find the solution in the first place. Just plug your solution in to the DE and see if you get equality! You got the same answer both ways, which is a good sign.

**alexmahone** · September 27th 2011, 07:50 AM

Originally Posted by Ackbeet

The beauty of DE's is it's often much easier to check your own solution than to find the solution in the first place. Just plug your solution in to the DE and see if you get equality! You got the same answer both ways, which is a good sign.

How would I differentiate $|\sin x|^{-1}$ and $|\sin x|^{-1/2}$ .

Also, are the modulus signs really necessary?

**Ackbeet** · September 27th 2011, 07:53 AM

Note that

$\frac{d}{dx}\,|x|=\frac{x}{|x|}=\text{sgn}(x),$

the signum function.

You can use the chain rule multiple times to get the final result.

**alexmahone** · September 27th 2011, 09:40 AM

Originally Posted by Ackbeet

Note that

$\frac{d}{dx}\,|x|=\frac{x}{|x|}=\text{sgn}(x),$

the signum function.

You can use the chain rule multiple times to get the final result.

$\frac{dy}{dx}=C^2*-1*|\sin x|^{-2}sgn(\sin x)\cos x$ $+2C*-\frac{1}{2}|\sin x|^{-3/2}sgn(\sin x)\cos x$

$=-C^2|\sin x|^{-2}sgn(\sin x)\cos x-C|\sin x|^{-3/2}sgn(\sin x)\cos x$

$\frac{\sqrt{y}-y}{\tan x}=\frac{-C^2|\sin x|^{-1}-C|\sin x|^{-1/2}}{\tan x}$

$=\frac{(-C^2|\sin x|^{-1}-C|\sin x|^{-1/2})\cos x}{\sin x}$

$=\frac{(-C^2|\sin x|^{-2}-C|\sin x|^{-3/2})\cos x}{\sin x|\sin x|^{-1}}$ (Multiplying numerator and denominator by $|\sin x|^{-1}$ .)

$=\frac{(-C^2|\sin x|^{-2}-C|\sin x|^{-3/2})\cos x}{sgn(\sin x)}$

$=(-C^2|\sin x|^{-2}-C|\sin x|^{-3/2})\cos xsgn(\sin x)$

$=-C^2|\sin x|^{-2}sgn(\sin x)\cos x-C|\sin x|^{-3/2}sgn(\sin x)\cos x$

$\frac{dy}{dx}=\frac{\sqrt{y}-y}{\tan x}$

QED

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