Question

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

Answer 1

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`

Answer 2

`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`.

Answer 3

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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