Two small diff. eq problems, think I am close though!

[ManyTimes]

Hey

1.)
y' = cosX * y + cosX
Y = | cosX * y + cosX + C //The | sign is "integral"
Y = -sinX * y - sinX + C

Is this completed? are there any further steps? Hmm...

2.)
Y’ + (1/x – 1) *y = x
Y’ = x – (1/x – 1)*y
Y' = 1/2x² - (1/x - 1) * y + C

I know this is wrong the part (1/x -1)*y is not correctly, i should use the core "backwards" (integral).
My try:
(1/x) - 1) :
1/x' = | ln(1/x) // Remember | is the "integral" sign
-1' = | -x
So I get:
Y(x) = 1/2x² - (ln(1/x) - x) * y + C

Have a feeling this is right, but yet not, cant be this easy and I think I should have e^ somewhere...

 

[wjm11]

y' = cosX * y + cosX
Y = | cosX * y + cosX + C //The | sign is "integral"
Y = -sinX * y - sinX + C

Is this completed?

You need to separate your variables first, then integrate.

y' = cosX * y + cosX
dy/dx = cosx(y + 1)
dy/(y + 1) = (cosx)(dx)

Now integrate.

 

[pitoten]

That is not how you do those. I'm not the best at diffiqus, but I see that the first problem is a "separable" differential equation. Go ahead and look that up...those ones aren't too bad.

Unfortunately, I do not think the second one is separable...but don't worry about that one yet.

 

[Deleted member 4993]

Hey

1.)
y' = cosX * y + cosX
Y = | cosX * y + cosX + C //The | sign is "integral"
Y = -sinX * y - sinX + C

Is this completed? are there any further steps? Hmm...

2.)
Y’ + (1/x – 1) *y = x

This one is of the form

y' + y*p(x) = g(x)

you need to multiply by integrating factor \(\displaystyle e^{\int p(x)dx}\)

now continue....

Y’ = x – (1/x – 1)*y
Y' = 1/2x² - (1/x - 1) * y + C

I know this is wrong the part (1/x -1)*y is not correctly, i should use the core "backwards" (integral).
My try:
(1/x) - 1) :
1/x' = | ln(1/x) // Remember | is the "integral" sign
-1' = | -x
So I get:
Y(x) = 1/2x² - (ln(1/x) - x) * y + C

Have a feeling this is right, but yet not, cant be this easy and I think I should have e^ somewhere...

 

[ManyTimes]

Edited the whole post; thanks for response so far!

1.)
y' = cosX * y + cosX
y'/y = cosX + cos X
ln(y) =e^(-sinX) + e^(-sinX)
y = e^(-sinX) + e^(-sinX) + C

2.)
y' + (1/x - 1) * y = X
y' + P(x)y = X
y' + P(x) = X/Y
y' * y = X - P(X)
1*y = x - p(X)
y = x - p(x)
y = 0.5x² - e^(p(x)) + C
y = 0.5x² - e^(ln(x) - x) + C

Now what do you say? I am impressed at least. But not for long I bet.

 

[Deleted member 4993]

Edited the whole post; thanks for response so far!

1.)
y' = cosX * y + cosX
y'/y = cosX + cos X
ln(y) =e^(-sinX) + e^(-sinX)
y = e^(-sinX) + e^(-sinX) + C

2.)
y' + (1/x - 1) * y = X
y' + P(x)y = X
y' + P(x) = X/Y <<<< How's that???? Why's that???? I don't think you are ready to do these problems
y' * y = X - P(X)
1*y = x - p(X)
y = x - p(x)
y = 0.5x² - e^(p(x)) + C
y = 0.5x² - e^(ln(x) - x) + C

Now what do you say? I am impressed at least. But not for long I bet.

 

[ManyTimes]

>>y' + P(x) = X/Y <<<< How's that???? Why's that???? I don't think you are ready to do these problems
I think I am!!! Or, well, I'm just studying math on my own, so... Think I am progressing just fine... Nice and easy!

Dividing by Y, to then move the free "P(x)" over to the other side just changing operator sign, to then get rid of it from the right side by multiplying it on the left?
Is that an illegal operation???

So the final answer is wrong? In both of the questions or only second one or none?

 

[Deleted member 4993]

>>y' + P(x) = X/Y <<<< How's that???? Why's that???? I don't think you are ready to do these problems
I think I am!!! Or, well, I'm just studying math on my own, so... Think I am progressing just fine... Nice and easy!

Dividing by Y,

How come then y' did not get divided by 'y'?

to then move the free "P(x)" over to the other side just changing operator sign, to then get rid of it from the right side by multiplying it on the left?
Is that an illegal operation???

So the final answer is wrong?

Yes - I have given you hints regarding method to solve the second one.

Others have given you hints regarding the first one.

In both of the questions or only second one or none?

 

[Deleted member 4993]

Hey

1.)
y' = cosX * y + cosX

y'/(1+y) = cos(x)

ln(1+y) = sin(x) + C[sub:2suutkb7]1[/sub:2suutkb7]

y = C*e[sup:2suutkb7]sin(x)[/sup:2suutkb7] - 1


Y = | cosX * y + cosX + C //The | sign is "integral"
Y = -sinX * y - sinX + C

Is this completed? are there any further steps? Hmm...

2.)
Y’ + (1/x – 1) *y = x

Multiplying by intigrating factor ? Integrating factor I = x/e[sup:2suutkb7]x[/sup:2suutkb7]

d/dx (y*x*e[sup:2suutkb7]-x[/sup:2suutkb7]) = x[sup:2suutkb7]2[/sup:2suutkb7]*e[sup:2suutkb7]-x[/sup:2suutkb7]

Now integrate both sides


Y’ = x – (1/x – 1)*y
Y' = 1/2x² - (1/x - 1) * y + C

I know this is wrong the part (1/x -1)*y is not correctly, i should use the core "backwards" (integral).
My try:
(1/x) - 1) :
1/x' = | ln(1/x) // Remember | is the "integral" sign
-1' = | -x
So I get:
Y(x) = 1/2x² - (ln(1/x) - x) * y + C

Have a feeling this is right, but yet not, cant be this easy and I think I should have e^ somewhere...