I'm stuck on the following differential equation question, I have to solve it and find a general solution in explicit form: dxdy=(1−y2)×sin(x)
My attempt is as follows, using the variable separable method: (1−y2)211dxdy=sin(x)
∫(1−y2)211dy=∫sin(x)dx
Here is where I got stuck integrating the LHS. Am I correct in integrating by substitution with y = sinx?
I tried it but the answer I got looked wrong, I got upto (for LHS): ∫sec2(x)dy
I'm not sure whether my integration by trig substitution is the problem or if this type of ODE doesn't work for the variable separation method, any insights would be appreciated.
I'm stuck on the following differential equation question, I have to solve it and find a general solution in explicit form: dxdy=(1−y2)×sin(x)
My attempt is as follows, using the variable separable method: (1−y2)211dxdy=sin(x)
∫(1−y2)211dy=∫sin(x)dx
Here is where I got stuck integrating the LHS. Am I correct in integrating by substitution with y = sinx?
I tried it but the answer I got looked wrong, I got upto (for LHS): ∫sec2(x)dy
I'm not sure whether my integration by trig substitution is the problem or if this type of ODE doesn't work for the variable separation method, any insights would be appreciated.
That is the correct substitution, however it seems like you forgot to square root the denominator after the sub
Also remember that if y=sinx then dy/dx = cosx and so dy=cosx dx meaning the numerator and denominator should end up cancelling out
Or just use the formula book since it's a standard integral
For the substitution, you will need to introduce another variable, so y=sinu is OK. Then recall that dudtanu=sec2u.
with ∫sec2(x)dy I have to integrate with respect to y, and I have an x value so I imagined it as a constant and having a 1 infront of it so turning that into a y with the constant following, I'm not sure what to do here. I can see what you mean and I would do that if the integral was with respect to x, do you end up with my answer when you try the substitution method?
No, this is what happens when you get too used to what "x" is meant to mean.
You want to substitute for dy in terms of dx, because your original integral is with respect to dy and you want to change it to dx. So dy = cos x dx.
Then ∫cosx1dy=∫cosx1⋅cosxdx=∫1dx
If I were you, I'd just avoid using x as a substitution variable and stick with u or something...
Ah, I see where I went wrong now thanks for pointing it out. You're right I'm too used to subbing the dy into the dx so I did it out of habit without thinking about it, I'll begin using u instead
I'm trying that now and I got the general solution in explicit form but I'm having trouble verifying it, I've attached my working. I need the sqrt(1-y^2) to be sin(C-cosx) to verify it but I can only get it in terms of cos, am I doing it right?
if my working is unclear i can rewrite it clearly including full question, it's all scattered atm, the question is in the first post
I'm trying that now and I got the general solution in explicit form but I'm having trouble verifying it, I've attached my working. I need the sqrt(1-y^2) to be sin(C-cosx) to verify it but I can only get it in terms of cos, am I doing it right?
if my working is unclear i can rewrite it clearly including full question, it's all scattered atm, the question is in the first post
Sorry for the late reply.
You differentiation is incorrect. The derivative of sin(C−cos(x)) is sinx×cos(C−cos(x)) and notsinx×sin(C−cos(x)) It should work then.
Let y=arcsin(x) ⇒x=siny dydx=cosy Using dxdydydx=1 dxdy=cosy1=1−sin2y1=1−x21
Technically, this is correct. But (I'm sure) the point Zacken is trying to make here is that you should be able to find this result by substitution.
To be clear, instead of using substitution, you can do this by recognition (which is kind of what you've done, otherwise you'd have no reason to "start" with arcsin x), or by looking it up in the formula book, and these are all fine valid methods.
But if someone tries to do this by substitution, and they can't, then they need to brush up on your substitution methods; someone telling them "it's easy to verify that arcsin x works" is not doing them a service in the long term.
with ∫sec2(x)dy I have to integrate with respect to y, and I have an x value so I imagined it as a constant and having a 1 infront of it so turning that into a y with the constant following, I'm not sure what to do here. I can see what you mean and I would do that if the integral was with respect to x, do you end up with my answer when you try the substitution method?
You have made the substitution y=sinx and your intuition is correct. However, you will need to use another variable not aready present, say u, instead of x. Remember the goal here. You're solving the ODE to find y as a function of x. When you write y=sinx you're actually claiming that this is the solution of the ODE!
You have made the substitution y=sinx and your intuition is correct. However, you will need to use another variable not aready present, say u, instead of x. Remember the goal here. You're solving the ODE to find y as a function of x. When you write y=sinx you're actually claiming that this is the solution of the ODE!
Ah I used x because the RHS was in terms of x so I thought it would be easier, it still works with x but for better presentation I should use u instead?
Ah I used x because the RHS was in terms of x so I thought it would be easier, it still works with x but for better presentation I should use u instead?
No, we use u instead of x because x is already present on the RHS. Never use x for substitutions if it is already present somewhere else!