Question

What is the solution of the following DIFFERENTIAL EQUATION for the initial value problem: y(0) = -2

Eqn: y' + (tanx)y = 2(cosx)^2

Answer 1

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Answer 2

I know integrating factors now. So I get the integrating factor to be:
secx.

Answer 3

After doing the simplification

Answer 4

But what I am getting when I solve for y is:
\[y = 2\cos\]

Answer 5

Oops

Answer 6

\[y = 2cosxsinx + C\]

Answer 7

I know:
\[\mu(x) = \frac{1}{cosx}\]

mu is the INTEGRATING FACTOR

Answer 8

I think that
\[
y=2 (\sin (x) \cos (x)-\cos
(x)
\]

Answer 9

This is after simplification

Answer 10

But how did you get that -cosx???

Answer 11

Ok this is what I did:

Answer 12

I start off with the equation:

y' + (tanx)y = 2(cosx)^2

I multiply both sides by the integrating factor. Let's call it a(x), not mu(x)


a(x)[y' + tanx*y] = 2(cosx)^2

Answer 13

Then I expanded:

Answer 14

a(x)*y' + a(x)*tanx*y = a(x)*(2cosx)^2 <-- I missed that a(x) on the RHS, sorry about that

Answer 15

Then, we can re-express a(x)tanx:

Recall:

a(x) = \[e^{\int\limits_{}^{}p(x)dx}\]

Answer 16

My p(x) is tanx

Answer 17

The problem is that on the RHS:

I have:
\[a(x)*2(cosx)^2 \]

Answer 18

Which will then simplify to:
\[secx*(2)(cosx)\]

Answer 19

cos^2x sorry

Answer 20

So then you end up getting
cosx on the RHS

Answer 21

Could you show me how you did your RHS

Answer 22

\[
-\cos(x)
\]
is the solution of the homogenous equation
\[
y'(x)+y(x) \tan (x)=0
\]

Answer 23

\[2*\sec(x)*(\cos(x)^2) = 2\cos(x)\]

Answer 24

I get the above?

Answer 25

\[2*\frac{1}{cosx}*cosx*cosx\] bow two cosx cancel to leave 2cosx or did I miss something

Answer 26

@eliassaab I found out how I lost that other cos(x)

Answer 27

Now, I am getting:

C = -2

Answer 28

So, the solution for this initial value problem is:


y = 2sinxcosx - 2cosx

Answer 29

I found out that I have to cross-multiply AFTER I put in that "C" to the other side

Answer 30

Would this be correct?

Answer 31

With the -2 in front of the cosx?

Answer 32

yes

Answer 33

Yes! Finally.

Answer 34

Finished.