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.

2 Answers
Jun 26, 2018

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 #

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_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)#