Given a solution
, we can attempt to find a solution of the form
. We have derivatives
Substituting into the ODE, we get
Setting
, we end up with the linear ODE
Multiplying both sides by
, we have
and noting that
we can write the ODE as
Integrating both sides with respect to
, we get
Now solve for
:
So you have
and given that
, the second term in
is already taken into account in the solution set, which means that
, i.e. any constant solution is in the solution set.
Answer:
Step-by-step explanation:
4-(3+1)^2-6/2+4
4-4^2-3+4
4-16-3+4
-12-3+4
-15+4
-11
-------------------------
7x^3+2x-5x^2+6+x+9+3x^3+12x^2
10x^3-5x^2+12x^2+2x+x+6+9
10x^3+7x^2+3x+15
-------------------------------
3x^2(4x^3-2x^2+7x-4)
12x^5-6x^4+21x^3-12x^2
--------------------------------------
6x^2+9x-7 when x=2
6(2)^2+9(2)-7
6(4)+18-7
24+18-7
42-7
35
------------------------------
3(x-y)^2+4y-6x+2
3(-1-3)^2+4(3)-6(-1)+2
3(-4)^2+12+6+2
3(16)+12+8
48+12+8
68
------------------------------
7x-5=23
7x=23+5
7x=28
x=28/7
x=4
---------------
7y-4+3y+6=9y+8
10y-4+6=9y+8
10y-9y+2=8
y+2=8
y=8-2=6
y=6
--------------------------
5(w-2)+6=3w-14
5w-10+6=3w-14
5w-4=3w-14
5w-3w-4=-14
2w-4=-14
2w=-14+4
2w=-10
w=-10/2
w=-5
----------------------
-4x+2<2x-6
-4x-2x+2<-6
-6x+2<-6
-6x<-6-2
-6x<-8
6x<8
x<8/6
x<4/3
x>4/3
Xz+y+1=z. given
y+1=z-xz. subtraction property of equality
y+1=z(1-x) distributive property of multiplication over addition/factoring
(1+y)/(1-x) = z division property of required, given x≠1
Answer
0.0555 …
Step-by-step explanation: