**Calc 2 Help Needed** Solve the initial value problem below:

(x^2 + 81)dy/dx = 3 y(9)=0

Question

**Calc 2 Help Needed** Solve the initial value problem below:

(x^2 + 81)dy/dx = 3   y(9)=0

amistre64 · Answer

calc2 has the easy separables

anonymous · Answer

I know this needs to be separated, but how?

amistre64 · Answer

$$(x^2+81)\frac{dy}{dx}=3$$
$$\frac{dx}{(x^2+81)}\left((x^2+81)\frac{dy}{dx}=3\right)$$
$$dy=\frac{3}{(x^2+81)}dx$$

amistre64 · Answer

then the left is a tan-1 type integration

anonymous · Answer

ok
so dy integrated is y

amistre64 · Answer

$$x=9tan(t)$$
$$dx=9sec^2(t)dt$$

$$dy=\frac{3}{((9tan(t))^2+81)}dx$$
$$dy=\frac{3*9sec^2(t)}{(81tan^2(t)+81)}dt$$
$$dy=\frac{27sec^2(t)}{81(tan^2(t)+1)}dt$$
$$dy=\frac{1}{39}\frac{sec^2(t)}{sec^2(t)}dt$$
then again, it might be pretty easy unless i messed it someplace

amistre64 · Answer

1/3 , not 1/39

anonymous · Answer

hmmm. I'm still lost. haha

anonymous · Answer

where did you get the x = 9tanx?

amistre64 · Answer

$$y=\frac13t$$

$$x = 9tan(t)$$
$$\frac x9 = tan(t)$$
$$tan^{-1}(\frac x9) = t$$
$$y=\frac13  tan^{-1}(\frac x9)+C$$
maybe lol

amistre64 · Answer

were did I get it?  from a trig substitution technique

amistre64 · Answer

i know if I had 81tan^2(t) + 81 I can manimpulate it into a 81sec^2(t) making the integration that much simpler

amistre64 · Answer

i got an A in typing class, honest :)

anonymous · Answer

haha.. well I'm horrible at calculus..

amistre64 · Answer

it just takes getting used to that you can actually adjust things in math.  We grow up thinking this thin is static and unforgiving

amistre64 · Answer

when I first learned this stuff im like; How do they expect us to know to do that? :)

anonymous · Answer

that's how I feel now. The class was relatively easy until 2 weeks ago.

anonymous · Answer

ok so doesn't there need to be a constant K on this problem? which I would plug back in with y(9) = 0?

amistre64 · Answer

i use C as an "c"onstant

amistre64 · Answer

we can call it M or W or $\Large \phi $ if we want to

anonymous · Answer

is the final answer just an integer?

amistre64 · Answer

by integer i take it yo mean some real number value; let see :)
$$y=\frac13  tan^{-1}(\frac x9)+C$$

$$0=\frac13  tan^{-1}(\frac 99)+C$$
$$0=\frac13  tan^{-1}(1)+C$$
$$0=\frac13  \frac{\pi}{4}+C$$
$$0=\frac{\pi}{12}+C$$
$$-\frac{\pi}{12}=C$$
therefore

$$y=\frac13  tan^{-1}(\frac x9)-\frac{\pi}{12}$$

amistre64 · Answer

the final answer is an equation that will solve the initial condition

anonymous · Answer

makes sense! somewhat! haha. thanks for your help, you're a dang genius.

amistre64 · Answer

youre welcome :)

anonymous · Answer

I have another one similiar if you want to help??

amistre64 · Answer

maybe in a minute or so if noone else takes it; i gotta rest form this one ...

anonymous · Answer

ok thanks!