[Sifr]

Solve

Cosx(dy/dx) + ysinx = cosxsin2x , y(0) = 1

Can I get a step by step guide as to how to solve this? I've kind of forgotten how to,

What I've done so far:
Get into form: dy/dx + P(x)y = Q(x)

So I have:
dy/dx + ytanx = sin2x

Integrating factor is: e^ Int(P(x))
= 1/cosx
(skipped the calculations here)

Where do I go from here? A step by step guide would be much appreciated!!

Thank you

[Intriguing Alias]

(Original post by Sifr)
Solve

Cosx(dy/dx) + ysinx = cosxsin2x , y(0) = 1

Can I get a step by step guide as to how to solve this? I've kind of forgotten how to,

What I've done so far:
Get into form: dy/dx + P(x)y = Q(x)

So I have:
dy/dx + ytanx = sin2x

Integrating factor is: e^ Int(P(x))
= 1/cosx
(skipped the calculations here)

Where do I go from here? A step by step guide would be much appreciated!!

Thank you

Integrating factor = I

[Sifr]

(Original post by hassi94)
Integrating factor = I

Why do we ignore the P(x)y ?

[bcrazy]

Once you have the integrating factor then really what you want do is to multiply both sides of the equation by it, in your case to get:
(1/cosx)*dy/dx + y*(1/cosx)*tanx = sin2x/cosx

The reason this is useful is because the LHS is now actually (y/cosx)' (ie. I(x)*y)

You can then integrate both sides with respect to x (the LHS will just be y/cosx, the RHS looks a bit nasty but I'll leave that to you!)

[Intriguing Alias]

(Original post by Sifr)
Why do we ignore the P(x)y ?

By using an integrating factor of e^int(p(x)) what we've actually done is made it possible to represent the LHS as one derivative.

If we multiply both sides by I, we get:

(1/cosx)(dy/dx) + y(1/cosx)tanx = sin2x/cosx

Which can be written as d/dx(y/cosx) = sin2x/cosx

You can use the product rule to find d/dx(y/cosx) if you want to show yourself that it works.

Then we just integrate both sides w.r.t. x

[porkstein]

(Original post by hassi94)
You can use the chain rule to find d/dx(y/cosx) if you want to show yourself that it works.

*Product rule.

[Intriguing Alias]

(Original post by porkstein)
*Product rule.

Woops I'm getting tired Thanks for that

[Sifr]

(Original post by bcrazy)
Once you have the integrating factor then really what you want do is to multiply both sides of the equation by it, in your case to get:
(1/cosx)*dy/dx + y*(1/cosx)*tanx = sin2x/cosx

The reason this is useful is because the LHS is now actually (y/cosx)' (ie. I(x)*y)

You can then integrate both sides with respect to x (the LHS will just be y/cosx, the RHS looks a bit nasty but I'll leave that to you!)

Why does the LHS become y/ cosx ?

[bcrazy]

(Original post by Sifr)
Why does the LHS become y/ cosx ?

Sorry- that was (y/cosx)' which was my notation for the differential of y/cosx with respect to x.
This is a case where it isn't immediately obvious it can be written in this form but you can differentiate this to check (and with integrating factor, getting the left hand side as the differential of something is always what you are trying to do and the method will always deliver this).

The reason for the method is below:

We want to change the equation of the form dy/dx + P(x)y=Q(x) into something of the form
d/dx (something*y) = ....
so that we can then integrate both sides, and to do this we will multiply both sides by an "integrating factor", call it R(x).

This would give us R(x)dy/dx + R(x)P(x)y = R(x)Q(x) [Equation 1]

Using the product rule, d/dx (R(x)y) = R(x)dy/dx + dR/dx * y [Equation 2]

Comparing the LHS of Equation 1 with the RHS of Equation 2, they will be the same if R(x)*P(x) = dR/dx
Using the method of separating variables, you can solve this to get
R(x)=e^Int(P(x))

So if you use this as your integrating factor then you will end up with a LHS that is equal to the LHS of equation 2.

There is a lot going on here and I have just discover how hard this is to explain without talking as well but hopefully that might make some sense as to why you do what you do. I think the Wiki page might have had an explanation which is probably more eloquent than mine so have a look at that as well!

[Intriguing Alias]

(Original post by bcrazy)
Sorry- that was (y/cosx)' which was my notation for the differential of y/cosx with respect to x.
This is a case where it isn't immediately obvious it can be written in this form but you can differentiate this to check (and with integrating factor, getting the left hand side as the differential of something is always what you are trying to do and the method will always deliver this).

The reason for the method is below:

We want to change the equation of the form dy/dx + P(x)y=Q(x) into something of the form
d/dx (something*y) = ....
so that we can then integrate both sides, and to do this we will multiply both sides by an "integrating factor", call it R(x).

This would give us R(x)dy/dx + R(x)P(x)y = R(x)Q(x) [Equation 1]

Using the product rule, d/dx (R(x)y) = R(x)dy/dx + dR/dx * y [Equation 2]

Comparing the LHS of Equation 1 with the RHS of Equation 2, they will be the same if R(x)*P(x) = dR/dx
Using the method of separating variables, you can solve this to get
R(x)=e^Int(P(x))

So if you use this as your integrating factor then you will end up with a LHS that is equal to the LHS of equation 2.

There is a lot going on here and I have just discover how hard this is to explain without talking as well but hopefully that might make some sense as to why you do what you do. I think the Wiki page might have had an explanation which is probably more eloquent than mine so have a look at that as well!

Think that was as good as explanation gets except the bolded is wrong. Assuming a typo

[bcrazy]

(Original post by hassi94)
Think that was as good as explanation gets except the bolded is wrong. Assuming a typo

Thanks, but am I being thick here- what is wrong about the bit in bold? The integrating factor should be e to the power of the integral of P(x)? No?

[Intriguing Alias]

(Original post by bcrazy)
Thanks, but am I being thick here- what is wrong about the bit in bold? The integrating factor should be e to the power of the integral of P(x)? No?

Wow sorry I thought that said ln as in log