Question

I need a little advice on this substitution problem:

Answer 1

So I have: \(\Large t^2 \frac{dy}{dt} +y^2=ty \)

I got it on linear form for a Bernoulli sub:
\(\Large \frac{dy}{dt} +\frac{y^2}{t^2}=\frac{y}{t} \)

\( u=y^{1-2} =y^{-1} ; du=-y^{-2} dy \)

Subbed in dy:
\(\Large -y^2 \frac{du}{dt} +\frac{y^2}{t^2} =\frac{y}{t} \)

Multiplied in u:
\(\Large -y \frac{du}{dt} +\frac{y}{t^2} = \frac{1}{t} \)

Got into linear again:
\(\Large \frac{du}{dt}-\frac{1}{t^2} =-\frac{1}{t} y^{-1} \)

Then from there it just gets weirder.. so I was wondering if I should have moved the \(ty\) to left and \(y^2 \) to the right in the beginning, what do you think?

Answer 2

try \(u = \frac{1}{y}\)

Answer 3

below is NOT standard form of bernouli - mistake : \[\frac{dy}{dt} +\frac{y^2}{t^2}=\frac{y}{t}\]

Answer 4

check again

Answer 5

It's \(\Large \frac{dy}{dx}+P(x)y=f(x) y^n \) right?

Answer 6

Yes

Answer 7

in your equation, power term and linear term got swapped

Answer 8

\[\frac{dy}{dt} -\frac{y}{t} = -\frac{y^2}{t^2}\]

substitute \(u = \frac{1}{y^{2-1}}\)

Answer 9

So in short, yes, I should have switched, thanks gane :P

Answer 10

np :) see if you can watch 10 minutes of below video
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-4-first-order-substitution-methods/

skip to 15th minute...

Answer 11

looks i have overlooked... your substitution is correct :)

Answer 12

after the substitution im getting this :
\[\dfrac{d\color{Red}{u}}{dt} + \frac{\color{Red}{u}}{t} = \frac{1}{t^2}\]

which is a linear eqn

Answer 13

which is exactly same as your last equation if you replace \(y^{-1}\) by \(u\)

Answer 14

I ended up getting \(\Large y^{-1}=\frac{ln ~t}{t} + c~t^{-1}\)

Is the right? Answer key looks totally different P:

Answer 15

answers should match
let me ask wolfram, one sec

Answer 16

The answer key gives \(\Large e^{\frac{t}{y}} =ct \)

Answer 17

both are right, lets see if we can convert your answer to the textbook form

Answer 18

\[y^{-1}=\frac{\ln ~t}{t} + c~t^{-1}\]

which is same as

\[\frac{1}{y}=\frac{\ln ~t}{t} +\frac{c}{t}\]

multiplying through out by \(t\)
\[\frac{t}{y}=\ln ~t +c\]

Answer 19

next take exponent both sides and define a new constant

Answer 20

\[\frac{t}{y}=\ln ~t +c\]

takinng exponent both sides


\[e^{\frac{t}{y}}=e^{\ln ~t +c}\]

Answer 21

\[e^{\frac{t}{y}}=e^{\ln ~t +c} = e^{\ln t}\cdot e^c = t\cdot e^c = tc_1\]

Answer 22

Yea, I realized how they did it after you flipped them around, thanks again, you're the best :D

Answer 23

<3