Show that (x^3)(dy/dx)+4=2x^2y can be reduced to dz/dx=-4/x^5 by using y=zx^2

hello132 said:
https://puu.sh/x8M1o/e3eda5a999.png

Using Chain Rule,
dz/dx = dz/dy x dy/dx

what i did was to make dy/dx the subject so i can satisfy the equation above, but i do not know how to find dz/dy
Click to expand...

Okay, so I'll begin by copying the text from the image for those who might not be so inclined to visit an outside link:

Show that the differential equation

\(\displaystyle x^3 \: \dfrac{dy}{dx} + 4 = 2x^2y\)

can be reduced to

\(\displaystyle \dfrac{dz}{dx} = -\dfrac{4}{x^5}\)

by means of the substitution \(\displaystyle y=zx^2\).

Hence, find y in terms of x, given that \(\displaystyle y = 6\) when \(\displaystyle x = 1\)
Click to expand...

Your working was to subtract 4 from both sides of the ODE and then divide by x3 to isolate \(\displaystyle \dfrac{dy}{dx}\). That leaves you with:

\(\displaystyle \dfrac{dy}{dx} = \dfrac{2x^2y - 4}{x^3}\)

\(\displaystyle \dfrac{dy}{dx} = \dfrac{2y}{x} - \dfrac{4}{x^3}\)

Edit: I had previously suggested a method that I now realize wouldn't work at all. My mistake.

At this point, you can rearrange the terms to get an ODE that can be solved by using what my textbook calls the "integrating factor method:"

\(\displaystyle \dfrac{dy}{dx} - \dfrac{2y}{x} = - \dfrac{4}{x^3}\)

Are you familiar with solving ODEs of the form:

\(\displaystyle \dfrac{dy}{dx} - f(x) \cdot y = g(x)\)

If so, where does applying those techniques lead you? If you're willing to skip ahead a bit and do the second part first, you should be able to get an expression for y in terms of x, using the given initial conditions. Then you can make the suggested substitution of y = zx2.
 
Last edited:
Thanks for the help.

Thanks for the help ksdhart2! Im able to solve the question.

Just an additional question, I am wondering what if we replace y with y2 instead.
We wouldn't be able to use the integrating factor method since it does not satisfy the equation as shown below.
[FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]y/[/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math-italic]f[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Math-italic]y[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math-italic]g[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT]

If that is the case how can we solve the question?

ksdhart2 said:
Okay, so I'll begin by copying the text from the image for those who might not be so inclined to visit an outside link:



Your working was to subtract 4 from both sides of the ODE and then divide by x3 to isolate \(\displaystyle \dfrac{dy}{dx}\). That leaves you with:

\(\displaystyle \dfrac{dy}{dx} = \dfrac{2x^2y - 4}{x^3}\)

\(\displaystyle \dfrac{dy}{dx} = \dfrac{2y}{x} - \dfrac{4}{x^3}\)

Edit: I had previously suggested a method that I now realize wouldn't work at all. My mistake.

At this point, you can rearrange the terms to get an ODE that can be solved by using what my textbook calls the "integrating factor method:"

\(\displaystyle \dfrac{dy}{dx} - \dfrac{2y}{x} = - \dfrac{4}{x^3}\)

Are you familiar with solving ODEs of the form:

\(\displaystyle \dfrac{dy}{dx} - f(x) \cdot y = g(x)\)

If so, where does applying those techniques lead you? If you're willing to skip ahead a bit and do the second part first, you should be able to get an expression for y in terms of x, using the given initial conditions. Then you can make the suggested substitution of y = zx2.
Click to expand...
 
hello132 said:
Thanks for the help ksdhart2! Im able to solve the question.

Just an additional question, I am wondering what if we replace y with y2 instead.
We wouldn't be able to use the integrating factor method since it does not satisfy the equation as shown below.
[FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]y/[/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math-italic]f[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Math-italic]y[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math-italic]g[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT]

If that is the case how can we solve the question?
Click to expand...

Glad I could be of help. Now, in your follow-up question, as I understand it you're asking about if the original equation had been given as:

\(\displaystyle x^3 \: \dfrac{dy}{dx} + 4 = 2x^2y^2\)

If that's not correct, please reply with any necessary corrections. Assuming this is the proper interpretation, then you're correct that you can't use the integrating factor method. Indeed, this equation proves very difficult to solve, and is far above my level of understanding. According to WolframAlpha, the solution involves modified Bessel functions of the first and second kind. If you're so inclined, you can read about them here (first kind)http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html and here (second kind)http://mathworld.wolfram.com/ModifiedBesselFunctionoftheSecondKind.html.
 
hello132 said:
Thanks for the help ksdhart2! Im able to solve the question.

Just an additional question, I am wondering what if we replace y with y2 instead.
We wouldn't be able to use the integrating factor method since it does not satisfy the equation as shown below.
[FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]y/[/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math-italic]f[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Math-italic]y[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math-italic]g[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT]

If that is the case how can we solve the question?
Click to expand...
That method is only valid for linear-first-order equation. Introduction of "y2" term violates linearity.

Solution of non-linear DFE is an advanced topic.
 
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