# Question Regarding Linear ODE-Trouble Using Integrating Factor

Consider the equation $$\frac{dy}{dx} + 5y = e^{2x}$$ One method of attack as far as i know is to multiply both sides by $e^{5x}$.This gives $$e^{5x}\frac{dy}{dx} + y5e^{5x} = e^{2x}e^{5x} = e^{7x}$$ We now find that the LHS is,in fact,the derivative of $ye^{5x}$. $$\therefore \frac{d}{dx}(ye^{5x}) = e^{7x}$$ Now what do i do?Integrate this way$$\int\frac{d}{dx}(ye^{5x}) = \int e^{7x} ?$$

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exactly! you have solved it in the first place, why post it :) –  Nikhil Bellarykar Dec 28 '11 at 8:44
Oh is it?I thought till the time you finish integrating the desired function it would still be considered unsolved.I don't know man. –  alok Dec 28 '11 at 8:53
I mean you have reached a stage where 'solving it further' is a mere formality. If you are confused wrt $dx$, just take it to RHS as Paul says and you're done. –  Nikhil Bellarykar Dec 28 '11 at 9:21

From $$\frac{d}{dx}(ye^{5x}) = e^{7x},$$ we get $$\int d(ye^{5x}) =\int e^{7x}dx,$$ which implies that $$ye^{5x}=\frac{1}{7}e^{7x}+C.$$

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Oh can we still can use $dx$ as a factor even though $\frac{d}{dx}(ye^{5x}) = e^{7x}$ is in $\frac{d}{dx}(f(x)) = f(x)$ form?I need some insight on this. –  alok Dec 28 '11 at 8:46
I am not sure...but that is what normally people do: If we have a differential equation $\frac{dy}{dx}+P(x)y=Q(x)$, after multiplying the integrating factor $I(x)=e^{\int P(x)dx}$, the differetial equation $I(x)\frac{dy}{dx}+I(x)P(x)y=Q(x)I(x)$ reduces to $\frac{d}{dx}(I(x)y)=I(x)Q(x)$. Then we can solve the equation by $\int d(I(x)y)=\int I(x)Q(x)dx$. –  Paul Dec 28 '11 at 8:52
yes exactly @Paul, author of this question can also use wolframalpha to make sure how to solve such probblems –  dato datuashvili Dec 28 '11 at 9:21

Yes, that's what you do.

Look at what you have: $${d\over dx} (ye^{5x}) = e^{7x}.$$ This is saying that $ye^{5x}$ is an antiderivative of $e^{7x}$. Thus, you can write $$\tag{1}ye^{5x} =\int e^{7x}\, dx;$$ after all, the indefinite integral of a function is its general antiderivative.

At this point, you then evaluate the integral in (1) and solve for $y$.

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