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15. Differential equations

Faculty of Economics and Business
University of Zagreb

15.1Introduction

When we were in elementary school, we were taught how to solve linear equations like

−2x+17=8x−13.-2x+17 = 8x-13.

The algorithm started with us moving all of the unknows to the one side, all of the knows to the other side and then explicitly solving the equation for xx. In what follows, we will introduce a new kind of an equation and then we will try and see which of the steps from the algorithm for solving linear equations can be used to solve this new type of an equation.

Now, our goal is to come up with an algorithm that we’ll be able to use for solving differential equations. As mentioned before, we will use the algorithm for solving linear equations and try to adapt its steps. So, in order to solve the linear equation given above, we have the following:

−2x+17=8x−13−2x−8x=−13−17[separate the unknows from the knowns]−10x=−30/ ⁣:(−10)[express the unknown]x=3\begin{split} -2x + 17 &= 8x - 13 \\ -2x - 8x &= -13 - 17 \quad [\text{\small separate the unknows from the knowns}] \\ -10 x &= - 30 / \colon (-10) \quad [\text{\small express the unknown}] \\ x &= 3 \end{split}

Take, for example, the differential equation x⋅y′=2x−1\displaystyle x \cdot y' = 2\sqrt{x} - 1. From the algorithm above, the first step in solving this differential equation would be to separate the unkowns from the knowns. Since we consider the function yy to be unknown, while the variable xx and all of the constants are know, we get the following:

y′=2x−1x.y' = \frac{2\sqrt{x} - 1}{x}.

Based on the algorithm for solving linear equations, the next step should be to express the unknown. In case of linear equations, we have expressed the unknown variable xx by dividing the whole equation by -10 since division is the opposite operation to multiplication. Similarly, in case of differential equations we have to integrate since integration is the opposite operation to differentiation. Hence, the algorithm for solving differential equations can be summarized as follows:

Lastly, let’s compare linear and differential equations side by side:

−2x+17=8x−13x⋅y′=2x−1type of equationlineardifferentialknownsconstantsconstants and expressions containing xunknownsxy\begin{array}{| c | c | c |} & -2x+17 = 8x - 13 & x \cdot y' = 2\sqrt{x} - 1 \\ \hline \text{\small type of equation} & \text{\small linear} & \text{\small differential} \\ \text{\small knowns} & \text{\small constants} & \text{\small constants and expressions containing } x \\ \text{\small unknowns} & x & y \end{array}

15.2Solving differential equations

Part 4. Integrals
14. Definite integrals and area