MHB Confirming what I learned, a follow up differential equation

[stripedcat]

With some help in this other thread

http://mathhelpboards.com/calculus-10/differiental-equation-question-particular-solutions-10864.html

I was able to see what I was doing wrong. Now I'm going to apply it to a different problem and see if I'm doing it right.

[math]dy/dx+(3/x)y=-16sin x^4[/math], y(1)=1

There should be an integration symbol before that 3/x, so its e^ integration symbol. I don't know how to format it...

[math]e^{ \int 3/x dx} = x^3[/math]

Multiply it out.

[math]x^3 dy/dx+3x^2y = -16x^3sin(x^4)[/math]

Apply the rule

[math]d/dx x^3y = -16x^3sin(x^4)[/math]

Integrate both sides dx

[math]x^3y = -4cos(x^4) + C[/math]

Isolate y

[math]y=(-4cos(x^4)+C)/(x^3)[/math]

Right so far?

Now for the y(1)=1

[math]1=(4cos(1)+C)/(1^3)[/math]

[math]1=(4cos(1)+C[/math]

[math]4cos(1) = ~2.1612[/math]

[math]1=2.1612+C[/math]

[math]C=-1.1612[/math]

[math]y=(4cos(x^4)+1.1612)/(x^3)[/math]

That right?

 

[DreamWeaver]

You can format your exponentiated integral like this:

e^{ \int f(x)\, dx}

$$e^{ \int f(x)\, dx }$$Or\exp \left( \int f(x)\, dx\ right)

$$\exp \left( \int f(x)\, dx\right)$$

 

[stripedcat]

There are several typos in the original question that I don't seem to be able to fix.

Should not be - in front of any of the 4cos for instance, and that should be - 1.1612

Still not sure if I should have that -1.1612 or the value for C I found.