Exact Differential Equation - Practice Questions & MCQ

Concepts Covered - 1

Exact Differential Equation

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

Illustration 1 :

$\text { Solution of the differential equation } 2 x y d x+\left(x^{2}+3 y^{2}\right) d y=0\;\text{is}$

Let us first separate terms containing only x with dx and terms containing only y with dy

$2xydx+x^2dy+3y^2dy=0$

Here first two terms have both x and y. We can make an observation that first two terms are the differentiation of x²y. Hence we can write this equation as

$d(x^2y) + 3y^2dy=0$

Integrating this, we get

$x^2y + y^3+c=0$

This is the solution of this equation

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

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

$\\\mathbf{6.\;\;\;}\mathit{\frac{y d x-x d y}{x y}=d\left[\log \left(\frac{x}{y}\right)\right]}\\\\$

Illustration 2 :

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

Observe that

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