# Why isn't dy/dx = 3x + 2y a linear differential equation?

## I've learned that linear differential equations are written in the form $\frac{\mathrm{dy}}{\mathrm{dx}} + P \left(x\right) y = Q \left(x\right)$. If you rewrote $\frac{\mathrm{dy}}{\mathrm{dx}} = 3 x + 2 y$ as $\frac{\mathrm{dy}}{\mathrm{dx}} - 2 y = 3 x$, wouldn't your $P \left(x\right)$ be $- 2$ and your $Q \left(x\right)$ be $3 x$, making this a linear differential equation? My teacher told me, however, that this is a nonlinear differential equation.

Jun 26, 2018

That is linear

#### Explanation:

With $y = y \left(x\right)$, the generalised linear DE is of form:

${a}_{o} \left(x\right) \textcolor{red}{y} + {a}_{1} \left(x\right) \textcolor{red}{y '} + {a}_{2} \left(x\right) \textcolor{red}{y ' '} + \ldots . + {a}_{n} \left(x\right) \textcolor{red}{{y}^{\left(n\right)}} = b \left(x\right)$

So what you say is true.

$\boldsymbol{2} y + \left(\boldsymbol{- 1}\right) y ' = - 3 x$

Or, re-arranging:

$y ' - 2 y = 3 x$

Jun 27, 2018

By definition, a DE is linear when, if ${y}_{1}$ and ${y}_{2}$ are solution of the homogeneous equation, then also any linear combination: $y = {c}_{1} {y}_{1} + {c}_{2} {y}_{2}$ is also a solution of the homogeneous equation.

Given the equation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 x + 2 y$

the corresponding homogeneous equation is:

$\frac{\mathrm{dy}}{\mathrm{dx}} - 2 y = 0$

Suppose ${y}_{1}$ and ${y}_{2}$ are solutions of the equation, then for any ${c}_{1} , {c}_{2}$ let:

$y = {c}_{1} {y}_{1} + {c}_{2} {y}_{2}$

Then:

$\frac{\mathrm{dy}}{\mathrm{dx}} - 2 y = {c}_{1} \frac{{\mathrm{dy}}_{1}}{\mathrm{dx}} + {c}_{2} \frac{{\mathrm{dy}}_{2}}{\mathrm{dx}} - 2 \left({c}_{1} {y}_{1} + {c}_{2} {y}_{2}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} - 2 y = {c}_{1} \left(\frac{{\mathrm{dy}}_{1}}{\mathrm{dx}} - 2 {y}_{1}\right) + {c}_{2} \left(\frac{{\mathrm{dy}}_{2}}{\mathrm{dx}} - 2 {y}_{2}\right)$

and as ${y}_{1} , {y}_{2}$ are solutions:

$\frac{\mathrm{dy}}{\mathrm{dx}} - 2 y = {c}_{1} \cdot 0 + {c}_{2} \cdot 0 = 0$

which proves the point.

You can also look at it in the following way: the equation is in the form:

$L \left(f\right) = 3 x$

where L(dot) is the differential operator:

$f \mapsto \frac{\mathrm{df}}{\mathrm{dx}} - 2 f$

and the operator L(dot) is linear as:

$L \left(\alpha f + \beta g\right) = \alpha L \left(f\right) + \beta L \left(g\right)$