## megannicole51 Group Title solve the differential equation. assume x,y,t>0 dy/dt=-yln(y/2). y(0)=1 8 months ago 8 months ago

1. megannicole51 Group Title

so i think im doing this right so far but I'm not really sure can someone work through the steps with me?

2. ddog1437 Group Title

are you taking calculus though?

3. megannicole51 Group Title

calc 2

4. megannicole51 Group Title

$\frac{ dy }{ -ylny } = \frac{ 1 }{ 2 }dt$

5. megannicole51 Group Title

ill show the steps i have so far and can u tell me if im right or not?

6. megannicole51 Group Title

ive never seen it done like that actually....my professor hasnt really done any examples so i am trying to figure it out from our book which isnt helping either

7. megannicole51 Group Title

@SithsAndGiggles

8. myininaya Group Title

$\frac{-1}{y \ln (\frac{y}{2})} dy= dt$

9. myininaya Group Title

He is doing separation of variables.

10. myininaya Group Title

I think he meant to divide -yln(y/2) on both sides.

11. myininaya Group Title

and then he multiplied both sides by dt

12. SithsAndGiggles Group Title

Thanks to @myininaya for pointing out my mistake: $$\color{blue}{\text{Originally Posted by}}$$ @SithsAndGiggles $\frac{dy}{dt}=-y\ln\left(\frac{y}{2}\right)\\ -\frac{1}{y\ln\left(\frac{y}{2}\right)}~dy=dt$ Integrate both sides: $-\int\frac{1}{y\ln\left(\frac{y}{2}\right)}~dy=\int dt$ First, a substitution: $$u=\dfrac{y}{2}$$, or $$2u=y$$, so that $$2du=dy$$: $-\int\frac{1}{2u\ln u}~(2~du)=\int dt\\ -\int\frac{1}{u \ln u}~du=\int dt$ Got everything so far? $$\color{blue}{\text{End of Quote}}$$