# Confused about linear vs non linear DE

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#### curious_mind

##### Full Member level 3 I quite understand linear equation such as y=mx+c, but I am confused about linear and non linear differential equations.

consider y=x^3, which is a non linear function. How is dy/dx=2x^2 a linear differential equation? Sorry if it is too elementary

#### pancho_hideboo Non linear.

#### albbg Your DE can be written as:

y' = 2*x^2.

It's of the first order because the maximum order of the derivative of "y" is 1.
It's not homogenous because it equates to something non zero
It's linear because, in general, the exponent of the variable "y" and all its derivatives is 1 (in this case you have only the first derivative to take into account)

#### pancho_hideboo Non linear.
Sorry, I’m wrong.

Assume two solutions, y1(x) and y2(x).

y3(x)=a1*y1(x)+a2*y2(x) also satisfies y3(x)=dy3(x)/dx

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#### pancho_hideboo dy(x)/dx=f(x)

dy1(x)/dx=f(x)
dy2(x)/dx=f(x)

d{a1*y1(x)+a2*y2(x)}=(a1+a2)*f(x)

#### curious_mind

##### Full Member level 3 Again Iam confused. if y=x^3 is non linear, then how y' = 2x^2 is linear by definition

#### pancho_hideboo #### albbg If I define the function y=ax+b, you know it's linear (or better linear affine since b is in general not equal to 0).
If I modify the coefficient "a" and "b" so that a=2t^2-3 and b=4t we can write:

y=(2t^2-3)x+4t

It is still linear (with respect to x): for every value of "t" I have a line on a graph. Then, to remain on the first order I can write, in general, the following DE in "y":

y'=ay+b

It is linear. Now I can modify the coefficient to be a=2x^2-3 and b=4x and I'll have:

y'=(2x^2-3)y+4x

as before, it's still linear.

In case of higher order it's the same: you can consider the case of multi variable functions y = a1*x1+a2*x2+a3*x3.... ==> y = a1y'+a2y''+a3y'''....

#### curious_mind

##### Full Member level 3 Once again, your example in #8 ,y=(2t^2-3)x+4t is clear as y changes wrt x and not t. But your second example,y'=(2x^2-3)y+4x, y prime changes with respect to x and not t in this case, it must be non linear

#### pancho_hideboo #### albbg Once again, your example in #8 ,y=(2t^2-3)x+4t is clear as y changes wrt x and not t. But your second example,y'=(2x^2-3)y+4x, y prime changes with respect to x and not t in this case, it must be non linear

OK, standard function: y is function of x, t is just a parameter. DE: y' is function of y, x is just a parameter.

#### pancho_hideboo Q(t)=C(t)*V(t) ; time variable capacitance

dQ(t)/dt=I(t)=dC(t)/dt*V(t) + C(t)*dV(t)/dt

Q(t) or I(t) ; any function of t
C(t) ; any function of t

Consider Q(t) or I(t) as stimulus and V(t) as response.

Obviously a relation between stimulus and response is linear.

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#### curious_mind

##### Full Member level 3 Revisiting the topic. I was always plotting the graph of LHS/RHS to check if it is linear or not. Setting aside the definition, dy/dx=2x^2 will result in parabola, when we plot the derivative against x. The example of i=c dv/dt is perfectly linear as current is directly proportional to rate of change wrt to time. In above example d^2y/dx^2= 4x would be perfectly linear

#### pancho_hideboo I was always plotting the graph of LHS/RHS to check if it is linear or not.
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In above example d^2y/dx^2= 4x would be perfectly linear
I can not understand what you want to mean at all.
Is there anyone who can understand except for you ?

dy(x)/dx=sin(x), dy(x)/dx=exp(x), etc. is linear DE.

#### wwfeldman from https://math.stackexchange.com/questions/414597/linear-vs-nonlinear-differential-equation

from the next to last response:

"Linear Differential equations are those in which the dependent variable and its derivatives appear only in first degree and not multiplied together"

and

quote
y′′′+y′′+y=ex
is linear The degree is irrelevant. What's important is "not multiplied together" – Scientifica Aug 28 '16 at 6:21
end of quote
note: in this quote, the x in ex is an exponent with base e (base of natural logarithms)

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