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$\int\frac{d}{dx}(ye^{5x}) = \int e^{7x} ?$
Question Regarding Linear ODE-Trouble Using Integrating Factor
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ordinary-differential-equations
integration
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0I 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. – 2011-12-28
2 Answers
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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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0yes exactly @Paul, author of this question can also use wolframalpha to make sure how to solve such probblems – 2011-12-28
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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 $y$.