Exact Differential Equation - Practice Questions & MCQ

Sometimes some differential equations can be solved using observation only. In such equations, we can get a differential of a function of x and y.

Illustration 1 :

Solution of the differential equation $2 x y d x+\left(x^2+3 y^2\right) d y=0$ is
Let us first separate terms containing only x with dx and terms containing only y with dy
$
2 x y d x+x^2 d y+3 y^2 d y=0
$

Here first two terms have both $x$ and $y$. We can make an observation that the first two terms are the differentiation of $x^2 y$. Hence we can write this equation as
$
d\left(x^2 y\right)+3 y^2 d y=0
$

Integrating this, we get
$
x^2 y+y^3+c=0
$

This is the solution of this equation

The presence of the following exact differentials should be observed in a given differential equation

1. $x d y+y d x=d(x y)$
2. $\quad x d x+y d y=\frac{1}{2} d\left(x^2+y^2\right)$
3. $\frac{x d y-y d x}{x^2}=d\left(\frac{y}{x}\right)$
4. $\frac{y d x-x d y}{y^2}=d\left(\frac{x}{y}\right)$
5. $\frac{x d y-y d x}{x y}=\frac{d y}{y}-\frac{d x}{x}=d\left[\log \left(\frac{y}{x}\right)\right]$
6. $\frac{y d x-x d y}{x y}=d\left[\log \left(\frac{x}{y}\right)\right]$

Illustration 2 :

$
\frac{x d y-y d x}{x^2+y^2}+e^x d x=0
$

Observe that
$
\frac{x d y-y d x}{x^2+y^2}=\frac{\frac{x d y-y d x}{x^2}}{1+\frac{y^2}{x^2}}=\frac{d\left(\frac{y}{x}\right)}{1+\left(\frac{y}{x}\right)^2}=d\left[\tan ^{-1}\left(\frac{y}{x}\right)\right]
$

So the equation is
$
d\left[\tan ^{-1} \frac{y}{x}\right]+e^x d x=0
$

Integrating
$
\tan ^{-1} \frac{y}{x}+e^x+c=0
$