# Differential Equation

• Feb 6th 2014, 03:56 AM
Deci
Differential Equation
Find a particular solution from the differential equation to the following

$\frac{dy}{dx} = \frac{ x^2 + 3y^2 }{2xy} ; y(1) = 2$

I can't separate the variables, and also when I use $y = vx$ the answers doesn't match the ans key which is $y = \sqrt{5x^3 - x^2}$

Same problem with this question

solve the DE : $\frac{dy}{dx} = \frac{ x^2 - y^2 }{xy}$

What method do we use for this type of DE?
• Feb 6th 2014, 04:21 AM
SlipEternal
Re: Differential Equation
Quote:

Originally Posted by Deci
Find a particular solution from the differential equation to the following

$\frac{dy}{dx} = \frac{ x^2 + 3y^2 }{2xy} ; y(1) = 2$

I can't separate the variables, and also when I use $y = vx$ the answers doesn't match the ans key which is $y = \sqrt{5x^3 - x^2}$

I get the answer key answer when I use $y = xv$. Post your work and maybe we can see where you are having trouble.

Quote:

Originally Posted by Deci
Same problem with this question

solve the DE : $\frac{dy}{dx} = \frac{ x^2 - y^2 }{xy}$

What method do we use for this type of DE?

Same method should work. Let $y = vx$.
• Feb 6th 2014, 03:56 PM
Prove It
Re: Differential Equation
Quote:

Originally Posted by Deci
Find a particular solution from the differential equation to the following

$\frac{dy}{dx} = \frac{ x^2 + 3y^2 }{2xy} ; y(1) = 2$

I can't separate the variables, and also when I use $y = vx$ the answers doesn't match the ans key which is $y = \sqrt{5x^3 - x^2}$

Same problem with this question

solve the DE : $\frac{dy}{dx} = \frac{ x^2 - y^2 }{xy}$

What method do we use for this type of DE?

See if you can write \displaystyle \begin{align*} \frac{dy}{dx} = f\left( \frac{y}{x} \right) \end{align*}. If so, then the substitution \displaystyle \begin{align*} u = \frac{y}{x} \end{align*} is appropriate.
• Feb 7th 2014, 07:50 AM
Deci
Re: Differential Equation
1)
$\frac{dy}{dx} = \frac{x^2 + 3y^2}{2xy}$
${2xy}\frac{dy}{dx} = {x^2 + 3y^2}$

$y = vx$
$\frac{dy}{dx} = v + x \frac{dv}{dx}$

$2x (vx) (v+x \frac{dv}{dx}) = x^2 + 3 (vx)^2$

$2v^2x^2 + 2vx^3 \frac{dv}{dx} = x^2 + 3v^2x^2$

$2vx^3 \frac{dv}{dx} = x^2 + 3v^2x^2 - 2v^2x^2}$

$2vx^3 \frac{dv}{dx} = x^2 (1 + v^2) ($

$2vx^3 \frac{dv}{dx} = x^2(1 + v^2)$

$x \frac{dv}{dx} = \frac{(1 + v^2)}{2v}$

$\frac{1}{x} dx = \frac{2v}{1+ v^2} dv$

$\frac{1}{x} dx = \frac{2v}{1+ v^2} dv$

integrate both sides

$ln|x| = ln | 1 + v^2|$

$x = 1 + v^2$

$v = \sqrt{1-x^2}$

$y = x \sqrt{1-x^2}$
• Feb 7th 2014, 08:31 AM
romsek
Re: Differential Equation
Quote:

Originally Posted by Deci
1)
$\frac{dy}{dx} = \frac{x^2 + 3y^2}{2xy}$
${2xy}\frac{dy}{dx} = {x^2 + 3y^2}$

$y = vx$

<snip>

integrate both sides

$ln|x| = ln | 1 + v^2|$

$x = 1 + v^2$

$v = \sqrt{1-x^2}$

$y = x \sqrt{1-x^2}$

a few things

you need to solve

$\left|x\right| = \left|1+v^2\right|$ so you'll have a second solution

$x=1+v^2 \Rightarrow v=\sqrt{x-1} not v = \sqrt{1-x^2}$ as you've written.

you've neglected the constant of integration. This needs to be solved for given the boundary condition.

Start at the point where you integrate both sides and rework it.
• Feb 8th 2014, 05:20 AM
Deci
Re: Differential Equation
Okay, I tried it again, please check whether my answer is correct or not

laTexis not working properly, so I type it like this

\frac{1}{x} dx = \frac{2v}{1+ v^2} dv

integrateboth sides

ln|x|= ln | 1 + v^2|

x= 1 + v^2

x= 1 + \frac({y}{x})^2 + C
y(1)= 2 --> 1 = 1 + 4 + C
C= -4
x= 1 + v^2 - 4
v= \sqrt{x+3}

\frac{y}{x}= \sqrt{x+3}
y= x \sqrt{x+3}
y= \sqrt{x^3 + 3x^2}
• Feb 8th 2014, 06:05 AM
SlipEternal
Re: Differential Equation
No, that is not correct. The constant of integration comes when you integrate, not when you exponentiate. On line 4, you write "integrate both sides". The next line should read:

ln|x| + C = ln|1+v^2|

Note that if you add a constant to both sides, you can subtract it from the right and the constant on the left "absorbs" the one from the right.

Now, when you exponentiate, you get:

e^(ln|x| + C) = 1+v^2
e^(ln|x|)e^C = 1+v^2
xe^C = 1+v^2

Now, I tend to write K for multiplicative constants. Let K = e^C:

Kx = 1+v^2

Now, Kx = 1 + (\frac{y}{x})^2, so y(1) = 2 --> K = 1+(\frac{2}{1})^2 = 5.

5x = 1+v^2 implies v = \pm \sqrt{5x-1}

y = xv = \pm x\sqrt{5x-1} = \pm \sqrt{5x^3-x^2}