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.

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