# Initial value problem

1. Jul 19, 2014

### JamesGoh

For the following problem

$\frac{dw}{d\theta}=\theta w^{2}sin(\theta^{2}), w(0)=1$

I am not able to obtain the solution

$w=\frac{2}{1+cos(\theta^{2})}$

Can anyone point out my mistake?

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

thanks

#### Attached Files:

File size:
34 KB
Views:
140
• ###### tute3q2cp2_smallres.jpg
File size:
28.6 KB
Views:
140
2. Jul 19, 2014

### SteamKing

Staff Emeritus
sin(θ$^{2}$) ≠ sin$^{2}$(θ)

Be careful where you put the exponents when using trig functions. It makes a difference for integrating and differentiating as well.

3. Jul 20, 2014

### JamesGoh

sorry wrong piece of paper.

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

#### Attached Files:

• ###### tute3q2c_small.jpg
File size:
38.2 KB
Views:
135
4. Jul 20, 2014

### verty

You forgot the arbitrary constant.

5. Aug 21, 2014

### rjr

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: Aug 21, 2014