#y=Pe^(ax) + Qe^(bx)# Show that #(d^2y)/dx^2-(a+b)dy/dx+aby=0#?

3 Answers
Feb 28, 2018

See below

Explanation:

Let #y=Pe^(ax)+Qe^(bx)#

Then

#aby=ab(Pe^(ax)+Qe^(bx))=abPe^(ax) + abQe^(bx)#

And # dy/dx=aPe^(ax)+bQe^(bx)#

#rArr (a+b)dy/dx = (a+b)(aPe^(ax)+bQe^(bx))= a^2Pe^(ax)+abQe^(bx)+abPe^(ax)+b^2Qe^(bx)#

And #(d^2y)/dx^2=a^2Pe^(ax)+b^2Qe^(bx)#

So

#(d^2y)/dx^2-(a+b)dy/dx+aby#
#=a^2Pe^(ax)+b^2Qe^(bx)-(a^2Pe^(ax)+abQe^(bx)+abPe^(ax)+b^2Qe^(bx)) +abPe^(ax) + abQe^(bx)=0, # as required. #square#

Feb 28, 2018

#color(red)(y=Pe^(ax)+Qe^(bx)rArry_2-(a+b)y_1+aby=0)#
Answer for above Question is given below.

Explanation:

We have,
#y=Pe^(ax)+Qe^(bx)#
#rArry=Pe^(ax)+Q/e^(-bx)#
#rArre^(-bx)y=Pe^(ax-bx)+Q#
#rArre^(-bx)y-Pe^(ax-bx)=Q#, where Q is constant
Diff.w.r.t.x,we gate,
#e^(-bx)y_1+y(e^(-bx))(-b)-Pe^(ax-bx)(a-b)=0#
Dividing both sides by, #e^(-bx)#
#y_1-by-Pe^(ax)(a-b)=0rArry_1-by-(P(a-b))/e^(-ax)=0#
#rArre^(-ax)y_1-be^(-ax)y-P(a-b)=0#,where P,a,b are constants.
Diff.w.r.t.x,we get,
#rArre^(-ax)y_2+y_1e^(-ax)(-a)-be^(-ax)y_1-bye^(-ax)(-a)=0#
Dividing both sides by, #e^(-ax)#
#rArry_2-ay_1-by_1-by(-a)=0#
#rArry_2-(a+b)y_1+aby=0#.

Feb 28, 2018

Kindly refer to a Proof in the Explanation.

Explanation:

Prerequisite : The elimination of #p and q# from the eqns. :

#px+qy+z=0, pl+qm+n=0, pu+qv+w=0# is,

#|(x,y,z),(l,m,n),(u,v,w)|=0#.

We have, #Pe^(ax)+qe^(bx)-y=0..........(1)#.

Diff.ing w.r.t. #x#, we get, #P*ae^(ax)+Q*be^(bx)-y'=0........(2)#.

Rediff.ing w.r.t. #x#, we get, #P*a^2e^(ax)+Q*b^2e^(bx)-y''=0...(3)#.

Eliminating #P and Q# from #(1),(2) and (3)#, we have,

#|(e^(ax),e^(bx),-y),(ae^(ax),be^(bx),-y'),(a^2e^(ax),b^2e^(bx),-y'')|=0#.

#:. e^(ax)e^(bx)|(1,1,-y),(a,b,-y'),(a^2,b^2,-y'')|=0#.

Since #e^(ax),e^(bx)!=0#, and applying #C_2-C_1#, we get,

#|(1,0,-y),(a,b-a,-y'),(a^2,b^2-a^2,-y'')|=0#.

#:. (b-a)|(1,0,-y),(a,1,-y'),(a^2,b+a,-y'')|=0#.

As #(b-a)ne0#, and applying #R_2-a*R_1, and R_3-a^2*R_1#,

# |(1,0,-y),(0,1,ay-y'),(0,b+a,a^2y-y'')|=0#.

Finally, expanding by #C_1#, we have,

#1*|(1,ay-y'),(b+a,a^2y-y'')|-0+0=0, i.e., #

#1(a^2y-y'')-(b+a)(ay-y')=0, or, #

# a^2y-y''-bay-a^2y+(b+a)y'=0#.

#rArr y''-(a+b)y'+aby=0# what is the same as to say that,

#(d^2y)/dx^2-(a+b)dy/dx+aby=0#.

Feel & Spread the Joy of Maths.!