Question

Find the particular solution of the differential equation:
$\dfrac{{dy}}{{dx}} = - \dfrac{{x + y\cos x}}{{1 + \sin x}}$ given that $y = 1$, when $x = 0$

Answer 1

We are given a differential equation, whose particular solution has to be found. We do this by separating the term with purely $x$ function from the differential and mix of $x$ and $y$ function. Then we find a differential equation of form such that we have to find the integrating factor. We solve it further ahead using the formulas to solve this problem.
Formula used: The solution of differential equation of form $\dfrac{{dy}}{{dx}} + yP(x) = Q(x)$ is given by
$y \times IF = \int {IF \times Q(x)dx}$, where $IF$ is the integrating factor given as, $IF = {e^{\int {P(x)dx} }}$.

Complete step-by-step solution:
WE are given the differential equation as,
$\dfrac{{dy}}{{dx}} = - \dfrac{{x + y\cos x}}{{1 + \sin x}}$
We now separate the term with purely $x$ function from the differential and mix of $x$ and $y$ function. $\Rightarrow \dfrac{{dy}}{{dx}} + \dfrac{{y\cos x}}{{1 + \sin x}} = - \dfrac{x}{{1 + \sin x}}$
This is of form,
$\dfrac{{dy}}{{dx}} + yP(x) = Q(x)$
So solve such differential equations, find the integrating factor $IF$ of $P(x)$. We find it as, $IF = {e^{\int {P(x)dx} }}$.Using this ,

$$\Rightarrow IF = {e^{\int { \dfrac{{\cos x}}{{1 + \sin x}}dx} }} \\\ \Rightarrow IF = {e^{\ln (1 + \sin x)}} \\\ \Rightarrow IF = 1 + \sin x \\\$$

Hence, we get the integrating factor as$1 + \sin x$. Now, we know that solution of a differential equation of kind $\dfrac{{dy}}{{dx}} + yP(x) = Q(x)$ is,
$y \times IF = \int {IF \times Q(x)dx}$
Using this solution we move ahead as,

$$\Rightarrow y \times (1 + \sin x) = \int {(1 + \sin x)} \left( { - \dfrac{x}{{1 + \sin x}}} \right)dx \\\ \Rightarrow y(1 + \sin x) = - \int {xdx} \\\ \Rightarrow y(1 + \sin x) = - \dfrac{{{x^2}}}{2} + c \\\ \Rightarrow y = - \dfrac{{{x^2}}}{{2(1 + \sin x)}} + \dfrac{c}{{1 + \sin x}}\,\,\,\, \to (1) \\\$$

Where $c$ is a constant.
Now to find the particular solution when we are given that $y = 1$, when $x = 0$,
Putting these values in equation $(1)$

$$\Rightarrow 1 = - \dfrac{{{0^2}}}{{2(1 + \sin 0)}} + \dfrac{c}{{1 + \sin 0}} \\\ \Rightarrow 1 = c \\\$$

We get the value of constant as $1$. Putting this in equation $(1)$ we get the particular solution as,
$y = - \dfrac{{{x^2}}}{{2(1 + \sin x)}} + \dfrac{1}{{1 + \sin x}}\,$

Note: We have to be careful in finding the form of the differential equation and then solving it as missing of finding its form may lead to more lengthy and complex calculations. When we once find the form of the given differential equation, the rest of the process becomes easy if we remember the formulas related to it.