Question

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

I've learned that linear differential equations are written in the form `dy/dx + P(x)y = Q(x)`.

If you rewrote `dy/dx = 3x + 2y` as
`dy/dx - 2y = 3x`,
wouldn't your `P(x)` be `-2` and your `Q(x)` be `3x`, making this a linear differential equation?

My teacher told me, however, that this is a nonlinear differential equation.

Answer 1

That is linear

Explanation:

With `y = y(x)`, the generalised linear DE is of form:

`a_o(x)color(red)(y)+a_1(x)color(red)(y')+a_{2}(x)color(red)(y'')+.... +a_n(x)color(red)(y^((n)))=b(x)`

So what you say is true.

`bb2 y +(bb(-1)) y' =- 3x`

Or, re-arranging:

`y' - 2y = 3x`

Answer 2

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_1y_1+c_2y_2` is also a solution of the homogeneous equation.

Given the equation:

`dy/dx =3x+2y`

the corresponding homogeneous equation is:

`dy/dx -2y = 0`

Suppose `y_1` and `y_2` are solutions of the equation, then for any `c_1,c_2` let:

`y= c_1y_1+c_2y_2`

Then:

`dy/dx -2y = c_1(dy_1)/dx +c_2(dy_2)/dx -2(c_1y_1+c_2y_2)`

`dy/dx -2y = c_1((dy_1)/dx-2y_1) +c_2((dy_2)/dx -2y_2)`

and as `y_1,y_2` are solutions:

`dy/dx -2y = c_1*0+c_2*0 = 0`

which proves the point.

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

`L(f) = 3x`

where `L(dot)` is the differential operator:

`f |-> (df)/dx -2f`

and the operator `L(dot)` is linear as:

`L(alphaf+betag) = alpha L(f)+ betaL(g)`