- #1
FaraDazed
- 347
- 2
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:
[itex]
\\ [/itex]
Part B: First I find the general solution
[tex]
\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}
[/tex]
Then the particular solution
[tex]
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
[/tex]
Last edited: