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Hello! I have been working on some differential equation homework in preparation for an upcoming exam. I understand that when trying to solve a differential equation that is not exact sometimes an integrating factor can be calculated and then multiplied to the whole question which would then make it possible to solve as an exact differential equation. I keep trying to solve for an integrating factor and then continue to solve the problem from there using the method i just described, but i am not getting the correct answer. I am running out of attempts on this online homework so I would really appreciate it if someone should help me solve this problem.

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    $\begingroup$ This is a linear ODE $\endgroup$ – Fakemistake Sep 15 '14 at 21:46
  • $\begingroup$ I am still not sure how to solve it …can you help me please? $\endgroup$ – user124539 Sep 17 '14 at 18:39
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So your differential equation is:

$$x+2y+\frac{dy}{dx}=0$$

Now, in order to use integrating factors, we need to analyze the coefficient of $y$

So we then have that:

$$e^{\int 2dx} = e^{2x}$$

Multiplying this to both sides:

$$e^{2x}\left(x+2y+\frac{dy}{dx}\right)=0$$

$$\left(e^{2x}y\right)' = -xe^{2x}$$

$$\int\left(e^{2x}y\right)' = \int-xe^{2x}$$

$$\int-xe^{2x} = \frac{e^{2x}}{4}(1-2x)$$

$$e^{2x}y = \frac{e^{2x}}{4}(1-2x) + C$$

$$y = \frac{1}{4}(1-2x) + \frac{C}{e^{2x}}$$

So one possible solution can be if $C = 1$:

$$y = \frac{1}{4}(1-2x) + \frac{1}{e^{2x}}$$

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In such situations you should presume that integrating factor is function of this or that. Let's assume that integrating factor is function of $x$ ($u(x)$). Then this should be met:

$$\frac{\partial}{\partial y}\left[u(x)P(x,y)\right]=\frac{\partial}{\partial x}\left[u(x)Q(x,y)\right] $$

By solving this, you should find:

$$u(x)=e^{\int F(x)dx} $$

In your case I find:

$$u(x)=e^{\int\frac{2x-1}{x}dx} $$

Therefore, integrating factor (as a function of $x$) is:

$$u(x)=\frac{e^{2x}}{x}$$

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