4 Solving exact equations

As we have seen, the differential equation $\frac{d}{dx}\left(yx\right)=3x^2$ has solution $y=x^2+C∕x$ . In the solution, $x^2$ is called the definite part and $C∕x$ is called the indefinite part (containing the arbitrary constant of integration). If we take the definite part of this solution, i.e. $y_{\text{d}}=x^2$ , then

$\frac{d}{dx}\left(y_{\text{d}}\cdot x\right)=\frac{d}{dx}\left(x^2\cdot x\right)=\frac{d}{dx}\left(x^3\right)=3x^2.$

Hence $y_{\text{d}}=x^2$ is a solution of the differential equation.

Now if we take the indefinite part of the solution i.e. $y_{\text{i}}=C∕x$ then

$\frac{d}{dx}\left(y_{\text{i}}\cdot x\right)=\frac{d}{dx}\left(\frac{C}{x}\cdot x\right)=\frac{d}{dx}\left(C\right)=0.$

It is always the case that the general solution of an exact equation is in two parts: a definite part $y_{\text{d}}\left(x\right)$ which is a solution of the differential equation and an indefinite part $y_{\text{i}}\left(x\right)$ which satisfies a simpler version of the differential equation in which the right-hand side is zero.

Task!

1. Solve the equation

   $\frac{d}{dx}\left(ycosx\right)=cosx$

2. Verify that the indefinite part of the solution satisfies the equation

   $\frac{d}{dx}\left(ycosx\right)=0.$

1. Integrate both sides of the first differential equation:

   Answer

   $ycosx=\int \; <!--nolimits\; -->cosxdx=sinx+C\text{leading to}y=tanx+Csecx$

2. Substitute for $y$ in the indefinite part (i.e. the part which contains the arbitrary constant) in the second differential equation:

   Answer

   The indefinite part of the solution is $y_{\text{i}}=Csecx$ and so $y_{\text{i}}cosx=C$ and

   $\frac{d}{dx}\left(y_{\text{i}}cosx\right)=\frac{d}{dx}\left(C\right)=0$