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

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

Question

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

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

Emma Aldridge · Accepted Answer

See below

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#

Samantha Adkison · Answer

Kindly refer to a Proof in the 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.!

Charles Anctil · Answer

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

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#.

Emily Ammerman · Answer

undefined To show that (d^2y)/dx^2 - (a+b)dy/dx + aby = 0 for the given function y = Pe^(ax) + Qe^(bx), we need to find the first and second derivatives of y with respect to x, and then substitute them into the differential equation.

First, find dy/dx:
dy/dx = aPe^(ax) + bQe^(bx).

Next, find d^2y/dx^2:
d^2y/dx^2 = a^2Pe^(ax) + b^2Qe^(bx).

Substitute these derivatives into the differential equation:
d^2y/dx^2 - (a+b)dy/dx + aby
= (a^2Pe^(ax) + b^2Qe^(bx)) - (a+b)(aPe^(ax) + bQe^(bx)) + ab(Pe^(ax) + Qe^(bx))
= a^2Pe^(ax) + b^2Qe^(bx) - a^2Pe^(ax) - abQe^(bx) - abPe^(ax) - b^2Qe^(bx) + abPe^(ax) + abQe^(bx)
= 0.

Therefore, (d^2y)/dx^2 - (a+b)dy/dx + aby = 0 for y = Pe^(ax) + Qe^(bx).