# Differential Equations - first order

1. Apr 9, 2014

1. The problem statement, all variables and given/known data

a function has the feature that at any point, the product of its gradient and the x-cordinate is equal to the square root of the y-cordinate multiplied by 5.

Part A: Write out a differential equation that describes the function
Part B: If the curve passes through the point x=7/3, y=2, find the particular solution the differential equation.

2. Relevant equations
n/a

3. The attempt at a solution
Right, I am extremely new to differential equations so go easy on me.

What I have done is below, I would be very grateful if someone could check my workings please. Thanks :)

p.s. sorry for the many steps, i like to do every little step

Part A:
$\\$

Part B: First I find the general solution
$$\begin{eqnarray} x (\frac{dy}{dx})&=&5\sqrt{y} \\ x \,\, dy &=& 5\sqrt{y} \,\,dx \\ dy&=&5\sqrt{y}x^{-1}\,\, dx \\ \frac{1}{\sqrt{y}} \,\,dy &=& 5x^{-1} \,\,dx \\ y^{-\frac{1}{2}} \,\,dy &=& 5x^{-1} \,\,dx \\ \frac{1}{5}y^{-\frac{1}{2}} \,\, dy&=&x^{-1} \,\,dx \\ \int\frac{1}{5}y^{-\frac{1}{2}} \,\, dy&=&\int x^{-1} \,\,dx \\ \frac{1}{5}\int y^{-\frac{1}{2}} \,\, dy&=&\int x^{-1} \,\,dx \\ \frac{1}{5}(2y^{\frac{1}{2}})&=&ln|x|+C \\ \frac{2}{5}\sqrt{y}&=&ln|x|+C \\ \end{eqnarray}$$

Then the particular solution
$$C=\frac{2}{5}\sqrt{y}-ln|x| \\ C=\frac{2}{5}\sqrt{2}-ln(\frac{7}{3}=-0.28 \\ \therefore \frac{2}{5}\sqrt{y}=ln|x|-0.28$$

Last edited: Apr 9, 2014
2. Apr 9, 2014

### vela

Staff Emeritus
I'd write $C = \frac 25 \sqrt 2 - \ln\frac 73$. Don't be so quick to plug things into a calculator.

And you need to solve for $y$. You should find
$$y = \left(\frac 52 \ln \frac{x}{7/3} + \sqrt 2\right)^2.$$ You can see by inspection that $y$ satisfies the initial condition, and you can differentiate it to verify it satisfies the differential equation.

3. Apr 10, 2014

Right ok, thanks. I didn't know I needed to solve for y.

Could I right that logarithm as $\ln \frac{3x}{7}$ or is that not allowed because its in the logarithm ?

Thanks :)

4. Apr 10, 2014

### vela

Staff Emeritus
Yeah, you can write the log that way. It just wrote it that way because it makes it really obvious that when x=7/3, the log vanishes, so y=2.

5. Apr 10, 2014