Differential Equation Help

[Alex874]

Hello! I am having trouble solving these two problems:

1) dy/dx=xy^2 given the initial condition y(3)=.2

2) dy/dx=x(y-1)^2 given the initial condition y(0)=-1

I keep getting the same wrong answer and do not know why. I have tried to find examples because seeing problems solved step by step really help me, but can’t seem to find anything for these types of equations. Any help would be greatly appreciated!

 

[Deleted member 4993]

Hello! I am having trouble solving these two problems:

1) dy/dx=xy^2 given the initial condition y(3)=.2

2) dy/dx=x(y-1)^2 given the initial condition y(0)=-1

I keep getting the same wrong answer and do not know why. I have tried to find examples because seeing problems solved step by step really help me, but can’t seem to find anything for these types of equations. Any help would be greatly appreciated!

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem

 

[Alex874]

Sorry. Here is what I have done. I think I am making a mistake after integrating as the answers in the textbook are completely different, I just don’t understand how to get there.

 

[JeffM]

[MATH]\int \dfrac{dy}{y^2} = \int y^{-2} \ dy = - y^{-1} + \text { a constant.}[/MATH]
Check your work.

[MATH]y = ce^{x^2/4} \implies \dfrac{dy}{dx} = \dfrac{2x}{4} * ce^{x^2/4} = \dfrac{x}{2} * y = \dfrac{xy}{2} \ne xy^2.[/MATH]
If you do the integration correctly, you get

[MATH]- y^{-1} = \dfrac{x^2}{2} + c_1 \implies \dfrac{1}{y} = \dfrac{c_2 - x^2}{2} \implies y = \dfrac {2}{c_2 - x^2}.[/MATH]
Now check.

[MATH]\dfrac{dy}{dx} = \dfrac{-2(-2x)}{(c_2 - x^2)^2} = x * \dfrac{4}{(c_2 - x^2)^2} = x \left ( \dfrac{2}{c_2 - x^2} \right )^2 = xy^2. \ \checkmark[/MATH]
Now finish it up.

Similar mistake in the second problem I believe though I have not reviewed that one carefully.

 

[HallsofIvy]

You have the same error in both problems. You seem to be under the impression that \(\displaystyle \int \frac{1}{f(x)}dx= ln(f(x))+ c\) for any function, f (or at least the two here). That is NOT true. What is true is that \(\displaystyle \int \frac{1}{x}dx\) and, by "substitution", \(\displaystyle \int \frac{1}{f(x)} f'(x)dx= ln(f(x))+ c\), but the "f'(x)" is important- it has to be in the integral to begin with.