Answer:
a-[tex]e^x,e^{-x}[/tex]
b-[tex]x^{-2}[/tex]
c-[tex]x^3[/tex]
Step-by-step explanation:
a.7 y''-7 y =0
Auxillary equation
[tex]D^2-1=0[/tex]
[tex](D-1)(D+1)=0[/tex]
D=1,-1
Then , the solution of given differential equation
[tex]y=e^x,y=e^{-x}[/tex]
2.[tex]7x^2y''+14xy'-14 y=0[/tex]
Y=[tex]y=x^3[/tex]
[tex]y=3x^2[/tex]
[tex]y''=6x[/tex]
Substitute in the given differential equation
[tex]42x^3+42x^3-14x^3\neq 0[/tex]
Hence, [tex]x^3 [/tex] is not a solution of given differential equation
[tex]e^x,e^{-x}[/tex] are also not a solution of given differential equation.
y=[tex]x^{-2}[/tex]
[tex]y'=-2x^{-3}[/tex]
[tex]y''=6x^{-4}[/tex]
Substitute the values in the differential equation
[tex]7x^2(6x^{-4})+14x(-2x^{-3})-14 x^{-2}[/tex]
=[tex]42x^{-2}-28x^{-2}-14x^{-2}=0[/tex]
Hence, [tex]x^{-2}[/tex] is a solution of given differential equation.
c.[tex]7x^2y''-42y=0[/tex]
[tex]y=x^3[/tex]
[tex]y'=3x^2[/tex]
[tex]y''=6x[/tex]
Substitute the values in the differential equation
[tex]42x^3-42x^3=0[/tex]
Hence, [tex]x^3[/tex] is a solution of given differential equation.
a-[tex]e^x,e^{-x}[/tex]
b-[tex]x^{-2}[/tex]
c-[tex]x^3[/tex]
Answer:
a)[tex]\boxed{7y''-7y=0\to y=e^{-x}}[/tex].
b)[tex]\boxed{7x^2y''+14xy'-14y=0\to y=x^{-2}}[/tex]
c)[tex]\boxed{7x^2y''-42y=0\to y=x^{-2}}[/tex]
[tex]\boxed{7x^2y''-42y=0\to y=x^{3}}[/tex]
Step-by-step explanation:
a) The given differential equation is: [tex]7y''-7y=0[/tex].
The characteristic equation is: [tex]7m^2-7=0[/tex]
This implies that: [tex]m=-1\:or\:\:m=1[/tex]
The auxiliary solution to this second order homogeneous differential equation is: [tex]y=Ae^{m_1x}+Be^{m_2x}[/tex]
Therefore any equation of the [tex]y=Ae^{x}+Be^{-x}[/tex] where A and B are constants is a solution [tex]7y''-7y=0[/tex].
[tex]\boxed{7y''-7y=0\to y=e^x}[/tex].
[tex]\boxed{7y''-7y=0\to y=e^{-x}}[/tex].
b) The given differential equation is: [tex]7x^2y''+14xy'-14y=0[/tex]
The characteristic equation is given by: [tex]am(m-1)+bm+c=0[/tex], where a=7, b=14 and c=-14
This implies that:
[tex]7m(m-1)+14m-14=0[/tex]
[tex]\implies m=-2\:or\:1[/tex]
The auxiliary equation is of the form: [tex]y=Ax^{m_1}+Bx^{m_2}[/tex] where A and B are constants.
Hence any equation of the form: [tex]y=Ax^{-2}+Bx[/tex] is a solution to
[tex]7x^2y''+14xy'-14y=0[/tex]
[tex]\boxed{7x^2y''+14xy'-14y=0\to y=x^{-2}}[/tex]
c) The given differential equation is: [tex]7x^2y''-42y=0[/tex]
The characteristic equation is given by: [tex]am(m-1)+bm+c=0[/tex], where a=7, b=0 and c=-42
This implies that:
[tex]7m(m-1)-42=0[/tex]
[tex]\implies m=-2\:or\:3[/tex]
The auxiliary equation is of the form: [tex]y=Ax^{m_1}+Bx^{m_2}[/tex] where A and B are constants.
Hence any equation of the form: [tex]y=Ax^{-2}+Bx^3[/tex] is a solution to
[tex]7x^2y''-42y=0[/tex]
[tex]\boxed{7x^2y''-42y=0\to y=x^{-2}}[/tex]
[tex]\boxed{7x^2y''-42y=0\to y=x^{3}}[/tex]
