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Given $dy/dx = 2y/(2x+1)$ and the initial condition $y(0)=e$ and $x>-1/2$, find the solution y.

I was able to separate the variables and integrate both sides respectively. I then tried to use the initial condition to find my Constant of integration but I get lost after that step. I don't know how i am supposed to isolate y from $ln(2y)$ on the left side of the equation.

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${{dy}\over y}={{2dx}\over{2x+1}}$ implies that $ln(|y|)=ln(|2x+1|)+c$, you deduce that $y=A(2x+1)$ and $A=e$.

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  • $\begingroup$ @Isham There is no square. Your integration is incorrect. $\endgroup$ – Dylan Apr 24 '18 at 13:37
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I post a full solution

$$dy/dx = 2y/(2x+1)$$ $$\frac {y'}{y}=\frac 2 {2x+1}$$ $$(\ln(y))'=\frac 2 {2x+1}$$ $$\ln(y)=2\int \frac {dx} {2x+1}$$ $$\ln(y)=\ln(2x+1)+K$$ Take the exponential $$y=e^{\ln(2x+1)+K}=Ke^{\ln(2x+1)}=K(e^{\ln(2x+1)})$$ $$y=K(2x+1)$$ $$y(0)=e \implies K=e$$ $$\boxed{y=e(2x+1)}$$

Note that $$\int \frac {dy}{2y}=\frac 12 \ln|y|$$ And that $$\int \frac {2dy}{y}=2\int \frac {dy}{y}=2 \ln|y|=\ln (y^2)$$ And more generaly for $\alpha$ constant: $$\int \alpha f(x) dx=\alpha\int f(x)dx$$

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    $\begingroup$ By the chain rule, $(\ln(2x+1))' = \frac{2}{2x+1} $ $\endgroup$ – Dylan Apr 24 '18 at 13:37
  • $\begingroup$ @Dylan Thanks corrected... $\endgroup$ – Isham Apr 24 '18 at 13:56

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