Question

Solve the differential equation:
$x + y\dfrac{{dy}}{{dx}} = \sec \left( {{x^2} + {y^2}} \right)$ . Also find the particular solution if $x = y = 0$.

Answer 1

Hint:
We can give substitution for the term inside the bracket. Then we can find its derivative and substitute in the given equation. Then we can integrate the equation by separating the variable. Then we can substitute back to get the required solution. Then we can apply the initial condition to find the value of the constant of integration. We can obtain the required solution by substituting the value of the constant of integration.

Complete step by step solution:
We have the equation $x + \dfrac{{dy}}{{dx}} = \sec \left( {{x^2} + {y^2}} \right)$ … (1)
We can give a substitution for the term inside the trigonometric function.
Let $t = {x^2} + {y^2}$ …. (2)
We can differentiate with respect to x.
 $ \Rightarrow \dfrac{{dt}}{{dx}} = 2x + 2y\dfrac{{dy}}{{dx}}$
On dividing throughout with 2, we get,
 $ \Rightarrow \dfrac{1}{2}\dfrac{{dt}}{{dx}} = x + \dfrac{{dy}}{{dx}}$ … (3)
On substituting equation (2) and (3) in (1), we get,
 $ \Rightarrow \dfrac{1}{2}\dfrac{{dt}}{{dx}} = \sec \left( t \right)$
On bringing the same variables to same side, we get,
 $ \Rightarrow \dfrac{{dt}}{{\sec t}} = 2dx$
We know that $\dfrac{1}{{\sec \theta }} = \cos \theta $
 $ \Rightarrow \cos tdt = 2dx$
On integrating on both sides, we get
 $ \Rightarrow \int {\cos tdt} = 2\int {dx} $
We know that $\int {\cos tdt} = \sin t$ and $\int {dx} = x$
 $ \Rightarrow \sin t = 2x + C$
On substituting equation (2), we get,
 $ \Rightarrow \sin \left( {{x^2} + {y^2}} \right) = 2x + C$
We are given the initial condition as $x = y = 0$. On applying the initial condition, we get,
 $ \Rightarrow \sin \left( {0 + 0} \right) = 2 \times 0 + C$
On simplification we get,
 $ \Rightarrow \sin 0 = 0 + C$
We know that $\sin 0 = 0$
 $ \Rightarrow 0 = C$
On substituting the value of C, we get,
 $ \Rightarrow \sin \left( {{x^2} + {y^2}} \right) = 2x + 0$
Hence, we have,
$ \Rightarrow \sin \left( {{x^2} + {y^2}} \right) = 2x$

So, the required general solution is $\sin \left( {{x^2} + {y^2}} \right) = 2x + C$ and the particular solution is $\sin \left( {{x^2} + {y^2}} \right) = 2x$.


Note:
We must give the substitution for the whole term inside the trigonometric function. While differentiating, we must consider both x and y as variables. As we are differentiating with respect to x, we must put the term [/dfrac{{dy}}{{dx}}] after differentiating y term. We must put the constant of integration after integrating. The general solution will have a constant term added to it. It represents a family of curves which is a solution of the differential equation. Particular solution is obtained by finding the value of the constant of integration. It is done by substituting the initial condition to the general solution.