Question

Solve the differential equation, $\cos (x + y)dy = dx$.

Answer 1

Hint: Try to reduce the number of variables by substituting a third variable and bring in relation between derivatives of new variables and above variables.

Complete step by step answer:
There are two variables $x,y$ and we have these inside a cosine as follows:
$\Rightarrow \cos (x + y)dy = dx$ …………………………………………….(1)
Simplest method of variable separation will not be helpful as the cosine term brings them in one term, thus inseparable. Instead we will remove this problem for a time till we simplify this expression. So, let us consider the term inside the cosine as a third variable. i.e. let us assume that $z=x+y$. Differentiating on both sides with respect to y, we get,
$\Rightarrow \dfrac{{dz}}{{dy}} = \dfrac{{dx}}{{dy}} + 1$
Therefore,
$\Rightarrow \dfrac{{dx}}{{dy}} = \dfrac{{dz}}{{dy}} - 1$ .............................................(2)
Substituting our assumption and (2) in (1), we get,
$\begin{gathered} \Rightarrow \cos (z) = \dfrac{{dz}}{{dy}} - 1 \\\ \Rightarrow \dfrac{{dz}}{{dy}} = \cos z + 1 \\\ \end{gathered}$
Now, we have reduced our problem to above form which is variable separable. But before that, we will use trigonometric relations to simplify it further. We know that and using this to the terms on right hand side we get,
$\Rightarrow \dfrac{{dz}}{{dy}} = 2{\cos ^2}\left( {\dfrac{z}{2}} \right)$
Now, we will proceed with a variable separable method with the last equation.
$\Rightarrow \dfrac{{dz}}{{{{\cos }^2}\left( {\dfrac{z}{2}} \right)}} = 2dy$
Integrating, on both sides without limits, we get
$\Rightarrow \int{{{\sec }^{2}}\left( \dfrac{z}{2} \right)dz=\int{2dy}}+C$, where is an integration constant.
We know that $\int {{{\sec }^2}\left( {at} \right)dt = } \dfrac{1}{a}\tan at + {C_1}$. Here, .
Thus solving the integration, we get,
$\Rightarrow 2\tan \left( {\dfrac{z}{2}} \right) = 2y + C$. Dividing both sides by 2, considering ${C_2} = \dfrac{C}{2}$ as new constant for integration and substituting back our assumption, $z = x + y$, we get
$\begin{gathered} \Rightarrow \tan \left( {\dfrac{{x + y}}{2}} \right) = y + {C_2} \\\ \Rightarrow \tan \left( {\dfrac{{x + y}}{2}} \right) - y = {C_2} \\\ \end{gathered}$
This is the solution.

Note: Here we substituted the derivative of x with respect to y. We can also proceed to the same solution if we consider the other way, i.e., derivative of y with respect to x.