Question

show that if y1 and y2 be the solutions of the equation dy/dx+Py=Q, where P and Q are functions of x alone, and y2=y1z , then z= 1+ ae^(-Q/y1)dx.a being arbitrary constant

Answer 1

differential equation....

Answer 2

yes

Answer 3

aww man I'm bad at proofs and I just went through a lot of math...umm you could ask phi, KingGeorge, OOOPS...

Answer 4

ok

Answer 5

z=1+ae^{(-\frac{Q}{y_1})}dx ?

Answer 6

\[z=1+ae^{(-\frac{Q}{y_1})}dx \]?

Answer 7

yes we have to prove it

Answer 8

can you help me in this?

Answer 9

I can get most of the way there, but the ending isn't working great... I'll show ya wht I have

Answer 10

yes pleaase

Answer 11

\[ \frac{\mathrm{d}y}{\mathrm{d}x}+Py = Q\]
y1 and y2 are solutions, so
\[ \frac{\mathrm{d}y_1}{\mathrm{d}x}+Py_1 = Q \quad \quad ; \quad
P= \Big (Q-\frac{\mathrm{d}y_1}{\mathrm{d}x}\Big) \frac{1}{y_1}\]
and
\[ \frac{\mathrm{d}y_2}{\mathrm{d}x}+Py_2 = Q \quad \quad ; \quad
P= \Big(Q-\frac{\mathrm{d}y_2}{\mathrm{d}x}\Big) \frac{1}{y_2}\]

so we can equate the Ps

\[\Big(Q-\frac{\mathrm{d}y_1}{\mathrm{d}x}\Big) \frac{1}{y_1}= \Big(Q-\frac{\mathrm{d}y_2}{\mathrm{d}x}\Big) \frac{1}{y_2}\]
using y2 = zy1 gives
\[\Big(Q-\frac{\mathrm{d}y_1}{\mathrm{d}x}\Big) \frac{1}{y_1}= \Big(Q-\frac{\mathrm{d}zy_1}{\mathrm{d}x}\Big) \frac{1}{zy_1}\]
the y1s cancel out, bring the z to the other side, and you can use the product rule on d zy1/dx
\[z\Big(Q-\frac{\mathrm{d}y_1}{\mathrm{d}x}\Big)= \Big(Q- \big(y_1\frac{\mathrm{d}z}{\mathrm{d}x}+z\frac{\mathrm{d}y_1}{\mathrm{d}x}\big)\Big)\]
\[zQ-z\frac{\mathrm{d}y_1}{\mathrm{d}x}= Q- y_1\frac{\mathrm{d}z}{\mathrm{d}x}-z\frac{\mathrm{d}y_1}{\mathrm{d}x}\]
the zdy1/dx's cancel out and you can move the Qs to one side
\[zQ-Q= - y_1\frac{\mathrm{d}z}{\mathrm{d}x}\]
\[-\frac{Q}{y_1}(z-1)= \frac{\mathrm{d}z}{\mathrm{d}x}\]
\[-\frac{Q}{y_1}\mathrm{d}x= \frac{1}{(z-1)}\mathrm{d}z\]

then you can integrate the right hand side to get ln(z-1), but I dunno how to keep the dx on the bottom on the left, or how to get the constant out front

\[-\int \frac{Q}{y_1}\mathrm{d}x= \int \frac{1}{(z-1)}\mathrm{d}z\]

\[-\int \frac{Q}{y_1}\mathrm{d}x=ln(z-1)+A\]

Answer 12

any thoughts? :P

Answer 13

let me see wait

Answer 14

It's just sooo close I think it's the right way to go, just dunno where to go :P

Answer 15

yes its right i m also stuck here

Answer 16

but thanx ,,

Answer 17

I mean you can get it to

\[-\int \frac{Q}{y_1}\mathrm{d}x=ln(z-1)+A\]
\[-\int \frac{Q}{y_1}\mathrm{d}x-A=ln(z-1)\]
\[e^{-\int \frac{Q}{y_1}\mathrm{d}x-A}=z-1\]
\[e^{-\int \frac{Q}{y_1}\mathrm{d}x}e^{-A}=z-1\]
\[ \quad \quad e^{-A} = a\]
\[z=1+ae^{-\int \frac{Q}{y_1}\mathrm{d}x}\]
but there's still that dumb integral :P

Answer 18

T_T

Answer 19

yes may be there is mistake in the question

Answer 20

I think that's the better answer ^_^

Answer 21

becoz this is correct answer

Answer 22

yes thanx a lot for your help

Answer 23

welcome ^_^ That was a fun one :D

Answer 24

:)