Initial Value Problem (complex example)

[King_Silver]

I know the method and can solve other initial value problems. This is the question given:
dy/dx + y^(-2) Sin(3x) = 0 for t > 0, with y(0) = 2.

I've brought the dy/dx and let it equal to the rest of the expression so it is now:
dy/dx = -y^-2 Sin(3x) , with y(0) = 2 (i.e. when x = 0, y = 2 )

The -y^-2 is obstructing me from integrating both sides and solving for constant C using the information given. What would I have to do to eliminate the -y^-2 entirely from the question so I can continue standard procedures?

 

[S.G. Janssens]

Maybe I'm overlooking something, but can't you just multiply the original equation by $y^2$?

 

[King_Silver]

Maybe I'm overlooking something, but can't you just multiply the original equation by $y^2$?

But if you multiply the original by y² then you will need to do the same to the left hand side as you have done to the right hand side, meaning the dy/dx would also be affected.

 

[SteamKing]

I know the method and can solve other initial value problems. This is the question given:
dy/dx + y^(-2) Sin(3x) = 0 for t > 0, with y(0) = 2.

It's not clear which is the dependent variable here. The equation suggests x, but there's this 't > 0' restriction. I'm confused.

I've brought the dy/dx and let it equal to the rest of the expression so it is now:
dy/dx = -y^-2 Sin(3x) , with y(0) = 2 (i.e. when x = 0, y = 2 )

It looks like you want to use separation of variables here. Do you know how to get all y and dy to one side of the equation, and all x and dx to the other?

The -y^-2 is obstructing me from integrating both sides and solving for constant C using the information given. What would I have to do to eliminate the -y^-2 entirely from the question so I can continue standard procedures?

If you separate the variables, everything should be smooth sailing after that.

 

[King_Silver]

It's not clear which is the dependent variable here. The equation suggests x, but there's this 't > 0' restriction. I'm confused.It looks like you want to use separation of variables here. Do you know how to get all y and dy to one side of the equation, and all x and dx to the other?If you separate the variables, everything should be smooth sailing after that.

What do you mean by that? all y to the left and all x terms to the right?

so like ∫ y^-2 dy = ∫Sin(3x) dx ? :)

 

[S.G. Janssens]

If you separate the variables, everything should be smooth sailing after that.

Indeed. Just allow yourself to do the usual abuse of differentials.

 

[SteamKing]

What do you mean by that? all y to the left and all x terms to the right?

so like ∫ y^-2 ∫dy = Sin(3x) dx ? :)

I think you have too many integral signs on one side of the equation, and not enough on t'other.

 

[King_Silver]

I think you have too many integral signs on one side of the equation, and not enough on t'other.

Sorry that was just an accident edited the post :) thanks! got it sorted now cheers! :D

 

[Samy_A]

What do you mean by that? all y to the left and all x terms to the right?

so like ∫ y^-2 dy = ∫Sin(3x) dx ? :)

Shouldn't that be ∫ y⁺² dy?

 

[King_Silver]

Shouldn't that be ∫ y⁺² dy?

And a -Sin(3x) I think, sorry my bad!
I'm sure it should be a -Sin(3x) as well.

 

[King_Silver]

integrated it.
y³/3 = 1/3 Cos(3x) + C, with y(0) = 2

so 2³/3 = 1/3 Cos(0) + C
2 2/3 = 1/3 + C

C = 2 1/3

That is what I got :)