Differential Equations - first order

In summary, the conversation discusses solving a differential equation for a function with a specific feature. The differential equation is found and the particular solution is determined by finding the general solution and then using the given point to find the specific value of the constant. The final solution is then verified to satisfy the differential equation.
  • #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]
 
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  • #2
FaraDazed said:

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]
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
vela said:
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 [itex]\ln \frac{3x}{7}[/itex] or is that not allowed because its in the logarithm ?

Thanks :)
 
  • #4
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
vela said:
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 :)
 

What is a first-order differential equation?

A first-order differential equation is an equation that involves a function and its derivative with respect to a single independent variable. It can be written in the form dy/dx = f(x,y).

What is the purpose of solving a first-order differential equation?

The purpose of solving a first-order differential equation is to find a function that satisfies the equation. This function can then be used to model and predict the behavior of a system over time.

How do you solve a first-order differential equation?

There are several methods for solving a first-order differential equation, including separation of variables, integrating factors, and using the method of undetermined coefficients. The most appropriate method depends on the specific equation and initial conditions.

What are the initial conditions in a first-order differential equation?

The initial conditions in a first-order differential equation refer to the specific values of the function and its derivative at a given point. These initial conditions are necessary for finding a unique solution to the equation.

What are some real-world applications of first-order differential equations?

First-order differential equations are used in a variety of fields, including physics, engineering, biology, and economics. They can be used to model population growth, chemical reactions, electrical circuits, and many other natural phenomena.

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