Question

(1-x)y''-4xy'+5y=cosx
state the order of the given ordinary
differential equation. Determine whether the equation is linear
or nonlinear by matching it with (6).

Answer 1

what do u know about order and degree ?

Answer 2

New dont know anything

Answer 3

order is simply the "highest derivative" in the equation :

\[\huge y^{\color{red}{''}} + y^2 = 0\]

order of above equation is \(\large \color{red}{2}\)

Answer 4

this is a completely free site, and ordinary differential equations are not so hard... there are only handful of methods to solve these. so did u get how to tell the order of a given equation ?

Answer 5

From the example you gave me i can see 2 which indicates second order, what about this (1-x)y''-4xy'+5y=cosx?? can we say second order as well

Answer 6

whats the `highest derivative` in this equation ?

Answer 7

(1-x) `y''`-4xy'+5y=cosx

Answer 8

since `y''` is the highest derivative, the order is `2`

Answer 9

So how do we determine whether the equation is linear
or nonlinear by matching it with (6).

Answer 10

this is similar to linear/non linear curves in coordinate plane

Answer 11

\[\large ax_1 + bx_2+cx_3+dx_4=c\]

this is a general form of linear equation in coordinate plane, right ?

Answer 12

notice that it has constant coefficients and there are not square or cube terms for the variables \(\large x_i\)

Answer 13

we can define "linear equation" in differential equations also similarly

Answer 14

but in differential equations,
the variables are `derivatives`, not the usual unknowns

Answer 15

for example, below are the variables :

\[\large y''', y'' , y', y\]

Answer 16

and the coefficients can be functions of x :
\[\large x^3, \sin x, \ln x, x^x\]

Answer 17

you can have ANY functions of \(\large x\) as coefficients

Answer 18

so here is a general form of a `linear differential equation` :

\[\large f(x) \color{Red}{y'''}+ g(x)\color{Red}{y''} + h(x)\color{Red}{y'} + i(x)\color{Red}{ y} = j(x)\]

Answer 19

this is lienar because none of the variables \(\large \color{red}{y'',y'',y',y}\) have squares or cubes

Answer 20

:) lets make above equation non-linear

Answer 21

below is NOT linear because the variable \(\large \color{Red}{y'''}\) has exponent 2 :

\[\large f(x) \color{Red}{(y''')^2}+ g(x)\color{Red}{y''} + h(x)\color{Red}{y'} + i(x)\color{Red}{ y} = j(x)\]

Answer 22

below is also NOT linear for the same reason :

\[\large f(x) \color{Red}{y'''}+ g(x)\color{Red}{y''} + h(x)\color{Red}{y'} + i(x)\color{Red}{ y^3} = j(x)\]

Answer 23

oh nice :) i am in india

Answer 24

can u tell if below is lienar ?

\[\large f(x) \color{Red}{y'''}+ g(x)\color{Red}{y''} + h(x)\color{Red}{y'} + i(x)\color{Red}{ \sin y} = j(x)\]

Answer 25

Has no exponential

Answer 26

So is Linear

Answer 27

but \(\large \color{Red}{\sin y}\) is NOT linear, right ?

Answer 28

so this is not linear.
in general, the varibles need to be linear

Answer 29

below is NOT linear either

\[\large f(x) \color{Red}{y'''}+ g(x)\color{Red}{y''} + h(x)\color{Red}{y'} + i(x)\color{Red}{ \ln y} = j(x)\]

Answer 30

cuz \(\large \color{Red}{\ln y}\) is NOT linear

Answer 31

Ok because of Ln y

Answer 32

exactly !

Answer 33

Alight Prof

Answer 34

Awesome

Answer 35

So a `linear differential equation` must have exact below form for \(\large \color{Red}{y}\) terms :

\[\large f(x) \color{Red}{y'''}+ g(x)\color{Red}{y''} + h(x)\color{Red}{y'} + i(x)\color{Red}{ y} = j(x)\]

Answer 36

good job

Answer 37

no messing with squares/cubes or other nonlinear functions

Answer 38

(sin ())y"' - (cos ())y'= 2 can this be 3 order non linear

Answer 39

you got the order correctly
but think again on linearity, is it really non linear in \(y\) ?

Answer 40

i dont see any `siny` or `y^2` terms in the equation

Answer 41

\[\large (\sin (x))\color{Red}{y'''} - (\cos (x))\color{Red}{y'}= 2\]

Answer 42

this is a perfect linear equation right ?

Answer 43

Great

Answer 44

Is is because the y is outside the bracket

Answer 45

all it matters is the dependent \(\large \color{Red}{y}\) terms and their derivatives.
NOT so much the independent \(x\) terms.

Answer 46

Woww great

Answer 47

\[\large (\sin (x))\color{Red}{y'''} - (\cos (x))\color{Red}{y'}= 2\]

compare this with below :

\[\large (f(x))\color{Red}{y'''} + (g(x))\color{Red}{y'}= h(x)\]

Answer 48

Clearly it is linear.
we don't care what f(x) or g(x) or h(x) are. we don't care about the independent terms in x.

Answer 49

Omg u are awesome

Answer 50

see if u can tell if below is linear/non linear :

\[\large x^2\color{Red}{y'''} + x\color{Red}{y'}= e^x\]

Answer 51

Linear

Answer 52

you got it !

Answer 53

In Problems 9 and 10, determine whether the given first-order
differential equation is linear in the indicated dependent
variable by matching it with the first differential equation
given in (7).
9. (y2- l)dx + x dy= O; in y; inx
10. udv + (v + uv -ue")du = O; in v; in u

Answer 54

I would say linear

Answer 55

Why not second order

Answer 56

lets see #9 :

\[\large (\color{red}{y^2}- 1)dx + x \color{red}{dy}= 0\]

Answer 57

Yes u wrote is correctly

Answer 58

this is NON-linear in \(\large \color{Red}{y}\) and the order is \(\large 1\)

Answer 59

however this is LINEAR in \(\large \color{red}{x}\) :

\[\large (y^2- 1)\color{Red}{dx }+ \color{red}{x}dy= 0\]

Answer 60

```
9. (y2- l)dx + x dy= O; in y; inx
```

non-linear in y
linear in x

Answer 61

does that look okay ?

Answer 62

You are a genius

Answer 63

lol no, you're a genius to get it correctly the first time itself :)

Answer 64

check ur message

Answer 65

please post next question in a new thread so that ur question comes in the top list and others jump in to assis you :)

you need to close this question... then only it allows u to open a new question

Answer 66

Ok thank u