Integrating a function with two variables

Question

If $y=f(x)$ and $\dfrac{dy}{dx} = \dfrac{x}{\cos(y)+1}$ find $y$ given that $y$ passes through $(0,0)$.

So we want to find

$$\int \frac{x}{\cos y+1}dx$$

Answer 1

Assuming $y$ is not a function of $x$, then $\cos y$ is invariant in the change of $x$. That is to say, if $y$ is not a function of $x$, if we change $x$ by a very small number, say $dx$, $y$ will not respond to the change. So, $\cos y$ will not respond. Thus, it is in this case, we can treat $\cos y$ to be an undetermined constant, just as we would $\cos a$, $\cos b$, etc.

If it is the case that $y$ is a function of $x$, we would have to use differential equations to solve the problem.

Since, it is the case that $y = f(x)$, then we solve this ODE by separation.

$$ \frac{dy}{dx} = \frac{x}{\cos y + 1} \iff (1 + \cos y)dy = xdx\\ \iff \int 1 + \cos y~dy = \int x dx \iff y + \sin y = \frac{1}{2}x^2 + c. $$

Now, we shall solve for $c$ by using the initial coniditions, when $x = 0, y = 0$. However, this clearly forces $c = 0$ because \begin{align} 0 + \sin 0 = \frac{1}{2}(0)^2 + c \iff 0 = c. \end{align}

Now solve for $y$ from here.

Answer 2

You cannot find $y$ using the integral you have written.

$(1+\cos y)dy =xdx$ gives $y+\sin\, y =x^{2}/2+C$ and the constant is $)$ because of the initial condition. SO the solution of the DE is given implicitly by $y+\sin y =x^{2}/2$. You cannot solve this explicitly.

Answer 3

Hint:

This is a Variable Separable Equation.

If $\frac{dy}{dx}=f(x)g(y)$ then $\int \frac{dy}{g(y)}=\int f(x)dx$

This then comes into the form $G(y)=F(x)+c$ where we apply $G^{-1}$ of both sides so our $y$ is alone on the LHS.

Additionally, use the conditions given to find the relevant $c$.

Try to follow this method with your question.

Answer 4

$$\int_0^y(\cos(y)+1)\,dy=\int_0^x x\,dx$$

gives $$\sin(y)+y=\frac{x^2}2.$$