Differential Equations - first order

[FaraDazed]

Homework Statement

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.

Homework Equations

n/a

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:
<br /> \\

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

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

 

[vela]

Homework Statement

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.

Homework Equations

n/a

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:
<br /> \\

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

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

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.

 

[FaraDazed]

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.

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 :)

 

[vela]

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.

 

[FaraDazed]

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.

Right I see, yeah that makes sense. Thanks :)