# Have I solved this ODE correctly?

In summary, this problem is about finding the y-coordinate of a point that is defined everywhere but for which the tangent at that point is undefined. The student has found a solution by using Weierstrauss substitution and integrating to get an equation in which te^x + xe^x - e^x + C = 0.f

## Homework Statement

y' + siny + xcosy + x = 0

## The Attempt at a Solution

well, I've used Weierstrauss substitutions: siny = 2t/(1+t^2) , cosy = (1-t^2)/(1+t^2) , dy = 2dt/(1+t^2) where t=tan(x/2).

2/(1+t^2)dt + 2t/(1+t^2)dx + x(1-t^2)/(1+t^2)dx + xdx = 0
2dt + 2tdx + x(1-t^2)dx + x(1+t^2)dx = 0
2dt + (2t + x(1-t^2) + x(1+t^2))dx = 0
2dt + (2t + x - xt^2 + x + xt^2)dx = 0
2dt + (2t + 2x)dx = 0
dt + (t+x)dx = 0

well, if I multiply e^x to both sides, it'll be an exact differential and after integrating I'll have:
te^x + xe^x - e^x + C. after substituting t= tan(x/2) we obtain:
e^x(tan(y/2) + x - 1) + C.

well, this is a problem that our professor gave in the last lecture for extra score in the final exam. so I want to be sure that I've solved it and my method is correct.
Is it correct if I substitute t=tan(x/2)? because the original problem is defined everywhere but after I use Weierstrauss substitutions then I should restrict x to the x's that are defined. (x=pi and -pi are obviously not defined now). Wouldn't that cause problems?

## Homework Statement

y' + siny + xcosy + x = 0

## The Attempt at a Solution

well, I've used Weierstrauss substitutions: siny = 2t/(1+t^2) , cosy = (1-t^2)/(1+t^2) , dy = 2dt/(1+t^2) where t=tan(x/2).

2/(1+t^2)dt + 2t/(1+t^2)dx + x(1-t^2)/(1+t^2)dx + xdx = 0
2dt + 2tdx + x(1-t^2)dx + x(1+t^2)dx = 0
2dt + (2t + x(1-t^2) + x(1+t^2))dx = 0
2dt + (2t + x - xt^2 + x + xt^2)dx = 0
2dt + (2t + 2x)dx = 0
dt + (t+x)dx = 0

well, if I multiply e^x to both sides, it'll be an exact differential and after integrating I'll have:
te^x + xe^x - e^x + C.
You should have an equation. Is it te^x + xe^x - e^x + C = 0?
after substituting t= tan(x/2) we obtain:
e^x(tan(y/2) + x - 1) + C.

well, this is a problem that our professor gave in the last lecture for extra score in the final exam. so I want to be sure that I've solved it and my method is correct.
Is it correct if I substitute t=tan(x/2)? because the original problem is defined everywhere but after I use Weierstrauss substitutions then I should restrict x to the x's that are defined. (x=pi and -pi are obviously not defined now). Wouldn't that cause problems?
You can check your solution yourself. Differentiate your final equation implicitly, and substitute into your differential equation.

You should have an equation. Is it te^x + xe^x - e^x + C = 0?
Yea. It's te^x + xe^x - e^x + C = 0. (I had written only the answer of the integration).

You can check your solution yourself. Differentiate your final equation implicitly, and substitute into your differential equation.

Of course, but that would be terribly tedious.

So it's OK for us to do the tedious work of checking, but not you?

So it's OK for us to do the tedious work of checking, but not you?

lol. I'm asking you to check if I've done it correctly. that means I'm asking others to see whether what I've done makes sense to them or not.

Who are you going to ask for a problem on a test? Checking your solution is a good habit to get into when you're solving differential equations.

Who are you going to ask for a problem on a test? Checking your solution is a good habit to get into when you're solving differential equations.

well, that's a good advice. I checked my solution and it worked. but there is one question remained to be answered. There are points like x=pi where tan(x/2) is not defined. those points certainly need to be taken care of I think. How should I treat those points?

well, I realized that I'll have to treat the points of the form y=2kπ + π separately (because for such points tan(y/2) is not defined and I'm missing some solutions of the ODE for that reason). if I plug y=2kπ+π into the equation I'll obtain:

y' + sin(2kπ+π) + xcos(2kπ+π) + x = 0
y' + 0 + x(-1) + x = 0 -> y' = 0 which means y = C.

Do I need to find what C is?

Any ideas?