Initial value problem

  • Thread starter JamesGoh
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  • #1
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For the following problem

[itex]\frac{dw}{d\theta}=\theta w^{2}sin(\theta^{2}), w(0)=1[/itex]

I am not able to obtain the solution

[itex]w=\frac{2}{1+cos(\theta^{2})}[/itex]

Can anyone point out my mistake?

I have attached my working out in a picture format below (may need to enlarge it)

thanks
 

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  • #2
SteamKing
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For the following problem

[itex]\frac{dw}{d\theta}=\theta w^{2}sin(\theta^{2}), w(0)=1[/itex]

I am not able to obtain the solution

[itex]w=\frac{2}{1+cos(\theta^{2})}[/itex]

Can anyone point out my mistake?

I have attached my working out in a picture format below (may need to enlarge it)

thanks
sin(θ[itex]^{2}[/itex]) ≠ sin[itex]^{2}[/itex](θ)

Be careful where you put the exponents when using trig functions. It makes a difference for integrating and differentiating as well.
 
  • #3
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sorry wrong piece of paper.

Please look at this new attachment and inform me of my error
 

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  • #4
verty
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You forgot the arbitrary constant.
 
  • #5
rjr
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nothing wrong with the u-substitution and integration, however on the right side toward the bottom of your page, one step reads:

1/w = 2/(cos(Θ2)) + C

What I suggest:

1). ALWAYS isolate the function variable first (in this case w(Θ)) before solving for the arbitrary constant (in this case C).

In other words - rather than what it reads now: 1/w(Θ) = ...

rearrange the equation such that it reads w(Θ) = ...

THEN solve with the initial value w(0) = 1

2). AlSO (and more importantly) keep in mind that any value multiplied by an arbitrary constant is still the value of the arbitrary constant (e.g. 2 * C = C)

So putting these two suggestions together, rewrite 1/w = 2/(cos(Θ2)) + C such that it reads 1/w(Θ) = .... then one ENTIRE fraction. Then flip the fractions on either side so that it reads w(Θ) = ...

THEN solve with the initial condition w(0) = 1.

hope this helps
 
Last edited:

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