Hi, I was wondering if anyone could help me with this problem. The differential equation y'=4y/x has a power function solution y=Ax^n if y(3)=162, determine the values of n and A. Thanks!
I'm sorry, ODE is short for ordinary differential equation. I assumed you were beginning a course in ordinary differential equations.
When you separate the variables, you typically get everything involving the dependent variable on the left and the independent variable on the right. For this ODE, you would then have:
$\displaystyle \frac{1}{y}\.dy=\frac{4}{x}\,dx$
Now, integrate...what do you get?
When you integrate, you will have:
$\displaystyle \int\frac{1}{y}\,dy=4\int\frac{1}{x}\,dx$
I am assuming you have been taught $\displaystyle \int\frac{1}{u}\,du=\ln|u|+C$. You need only write the constant of integration on one side, and it is typically written on the right side with the independent variable. It can be thought of as the composite constant from both integrations.
See if you can write the next step, and from there we will use the properties of logs to get the desired form.
Hmm...if you have not seen that rule of integration, then perhaps you are to just take the given solution:
(1) $\displaystyle y=Ax^n$
and then differentiate it with respect to $\displaystyle x$ and write the resulting differential equation in the given form. Using this and the point on the solution curve (3,162), you will then have two equations and two unknowns from which you can determine $\displaystyle A$ and $\displaystyle n$.
What do you get when you differentiate (1) with respect to $\displaystyle x$?
I apologize for the confusion, I assumed you were to solve the ODE and show that the solution is of the given form.
First, compute the derivative
$\displaystyle y = Ax^{n} \Rightarrow y' = Anx^{n-1}$
Substitute $\displaystyle y$ and $\displaystyle y'$ in the differential equation:
$\displaystyle Anx^{n-1} = \frac{4Ax^n}{x}$
$\displaystyle \Leftrightarrow n=4$
At this point we can use that $\displaystyle y(3)=162$ thus
$\displaystyle y(3)=162 = 81A \Leftrightarrow A = 2$
Therefore the solution is:
$\displaystyle y=2x^4$
@ chyksparks333
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-Dan
I'm a bit late on this but note that the problem does NOT ask you to solve the equation. You are given that the solutions is $\displaystyle y= Ax^n$ for some A and n and you are asked to find A and n. Siron's method- find the derivative of y, then put that and y into the equation, is the correct idea.