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I have to solve the following ODE $$y'=\frac{xe^{y}}{4x-1}$$ with $y(0)=0$. From the equation, I get $$\int e^{-y}dy=\int\frac{x}{4x-1}dx$$ In the second integral, if $u=4x-1\Rightarrow x=(u+1)/4$ and $du=4dx$; hence $$\int\frac{x}{4x-1}dx=\frac{1}{16}(4x-1+\ln|4x-1|)$$ but, if I evaluate the second integral using long division: $$\frac{x}{4x-1}=\frac{1}{4}\left(1+\frac{1}{4x-1}\right)$$ my final answer will be $$\int\frac{x}{4x-1}dx=\frac{1}{4}\left(x+\frac{1}{4}\ln|4x-1|\right)=\frac{1}{16}(4x+\ln|4x-1|)$$

so, I have two possible solutions: $$-e^{-y}=\frac{1}{16}(4x-1+\ln|4x-1|)+c$$ or $$-e^{-y}=\frac{1}{16}(4x+\ln|4x-1|)+c$$ with $c\in\mathbb{R}$. Using the fact $y(0)=0$, in the first equation $$c=-\frac{15}{16}$$ and in the second one $$c=-1$$ Which one is correct and why?

mvfs314
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1 Answers1

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You got the solution of the given differential equation as $$-e^{-y}=\frac{1}{16}(4x-1+\ln|4x-1|)+c\tag1$$where $c$ is an arbitrary constant and it can be written as $$-e^{-y}=\frac{1}{16}(4x-1+\ln|4x-1|)+c\\\implies -e^{-y}=\frac{1}{16}(4x+\ln|4x-1|)+c'\tag2$$where $~c'=c-1/16~$ is an arbitrary constant.

Now putting the initial condition $~y(0)=0~,$ from $(1)$ you get $~c=-15/16~$ and from $(2)$ you get $~c'=-1~.$

Now putting $~c=-15/16~$ in equation $(1)$, you get $$-e^{-y}=\frac{1}{16}(4x-1+\ln|4x-1|)-\frac{15}{16}\\\implies -e^{-y}=\frac{1}{16}(4x+\ln|4x-1|)-1\tag3$$ Again putting $~c'=-1~$ in equation $(2)$, you get $$-e^{-y}=\frac{1}{16}(4x+\ln|4x-1|)-1$$which is same as equation $(3)$.

So from here you can find that the particular solution of the given differential equation is same in both cases and it is independent of the choice of the arbitrary constant.

Therefore your both answers are correct and you are free to choose any one of them.

nmasanta
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