Question

Solve the DE:
\[\frac{\text dy}{\text dx}=\frac{2x+3y+4}{3x+y-1}\]

Answer 1

\[\text{let } y=Y+q,\qquad x=X+p\]
\[\frac{\text dY}{\text dX}=\frac{2(X+p)+3(Y+q)+4}{3(X+p)+(Y+q)-1}\]
\[\frac{\text dY}{\text dX}=\frac{2X+3Y+(2p+3q+4)}{3X+Y+(3p+q-1)}\]
\[2p+3q+4=0\qquad3p+q-1=0\]\[\qquad\qquad\qquad\qquad q=1-3p\]\[2p+3(1-3p)+4=0\qquad\qquad\qquad\qquad\]\[-7p+7=0\qquad\qquad\]
\[p=1\qquad\qquad q=-2\]\[y+2=Y,\qquad x-1=X\]

Answer 2

So this is the tutorial right ?

Answer 3

Uncle Rhaukus you knew the answer ?

Answer 4

\[\frac{\text dY}{\text dX}=\frac{2X+3Y}{3X+Y}\]\[\frac{\text dY}{\text dX}=\frac{2+3\frac YX}{3+\frac YX}\]\[\text{let } \frac YX=V\]\[Y= VX\]\[\frac{\text dY}{\text dX}=V+\frac{\text dV}{\text dX}\]
\[V+\frac{\text dV}{\text dX}=\frac{2+3V}{3+V}\]\[\frac{\text dV}{\text dX}=\frac{2+3V}{3+V}-V\]\[\frac{\text dV}{\text dX}=\frac{2+3V}{3+V}-V\frac{3+V}{3+V}\]\[\frac{\text dV}{\text dX}=\frac{2+3V-3V-V^2}{3+V}\]\[\frac{\text dV}{\text dX}=\frac{2-V^2}{3+V}\]\[\frac{3+V}{2-V^2}\text dV=\text dX\]

Answer 5

should i say gr8 work ? or should i say where r u having problem .. sorry i dont know much about this but seems u knw every thing about this :)

Answer 6

\[\int\frac{3+V}{2-V^2}\text dV=\int\text dX\]\[\int\frac{3}{2-V^2}+\frac{V}{2-V^2}\text dV=X+c\]\[-\frac {3}{2\times\sqrt 2}\ln\left|\frac{\sqrt 2+V}{\sqrt 2-V}\right|-\frac 12\ln\left|\sqrt {2}^2 -V^2\right|=X+c\]\[\frac {3}{\sqrt 2}\ln\left|\frac{\sqrt 2+V}{\sqrt 2-V}\right|-\ln\left|2-V^2\right|=-2X+c_1\]\[\frac {3}{\sqrt 2}\ln\left|\frac{\sqrt 2+\frac YX}{\sqrt 2-\frac YX}\right|-\ln\left|2-\left(\frac YX\right)^2\right|=-2X+c_1\]\[\frac {3}{\sqrt 2}\ln\left|\frac{X\sqrt 2+Y}{X\sqrt 2-Y}\right|-\ln\left|2\left(\frac {X}{X}\right)^2-\left(\frac YX\right)^2\right|=-2X+c_1\]\[\frac {3}{\sqrt 2}\ln\left|\frac{(x-1)\sqrt 2+(y+2)}{(x-1)\sqrt 2-(y+2)}\right|-\ln\left|\left(\frac {2(x-1)^2-(y+2)^2}{(x-1)^2}\right)^2\right|=-2(x-1)+c_1\]\[\frac {3}{\sqrt 2}\ln\left|\frac{\sqrt 2 x+y+2-\sqrt 2}{\sqrt2x-y+2-\sqrt 2}\right|-\ln\left|\frac {2x^2-4x+2-y^2-4y-4}{x^2-2x+1}\right|=2-2x+c_1\]\[\frac {3}{\sqrt 2}\ln\left|\frac{\sqrt 2 x+y+2-\sqrt 2}{\sqrt2x-y+2-\sqrt 2}\right|-\ln\left|\frac {2x^2-y^2-4x-4y-2}{x^2-2x+1}\right|=2-2x+c_2\]

Answer 7

∫3+V2−V2dV=∫dX
∫32−V2+V2−V2dV=X+c
−32×2√ln∣∣∣2√+V2√−V∣∣∣−12ln∣∣2√2−V2∣∣=X+c
32√ln∣∣∣2√+V2√−V∣∣∣−ln∣∣2−V2∣∣=−2X+c1
32√ln∣∣∣∣2√+YX2√−YX∣∣∣∣−ln∣∣∣2−(YX)2∣∣∣=−2X+c1
32√ln∣∣∣X2√+YX2√−Y∣∣∣−ln∣∣∣2(XX)2−(YX)2∣∣∣=−2X+c1
32√ln∣∣∣(x−1)2√+(y+2)(x−1)2√−(y+2)∣∣∣−ln∣∣∣∣(2(x−1)2−(y+2)2(x−1)2)2∣∣∣∣=−2(x−1)+c1
32√ln∣∣∣2√x+y+2−2√2√x−y+2−2√∣∣∣−ln∣∣∣2x2−4x+2−y2−4y−4x2−2x+1∣∣∣=2−2x+c1
32√ln∣∣∣2√x+y+2−2√2√x−y+2−2√∣∣∣−ln∣∣∣2x2−y2−4x−4y−2x2−2x+1∣∣∣=2−2x+c2

must say gr8 work @UnkleRhaukus

Answer 8

typing this all is such a gr8 work ...

Answer 9

that above is my work.

the answer in the back of the book is
____________________________________________________________
\[\ln\left|{2x^2-y^2-4x-4y-2}\right|=\frac {3}{\sqrt 2}\ln\left|\frac{\sqrt 2 x+y+2-\sqrt 2}{\sqrt2x-y+2-\sqrt 2}\right|\] ____________________________________________________________
my problem is that i can see how to get this answer

Answer 10

above my head

Answer 11

same here @vishweshshrimali5

Answer 12

which step is over your head?

Answer 13

Unkle Rhaukus I am new to calculus and that also D.E. and then your difficult steps.

Answer 14

it is scary looking stuff,

Answer 15

well i found one mistake so far in ur work

Answer 16

oh good.

Answer 17

\[Y=VX\]\[\frac{dY}{dX}=V+X \frac{dV}{dX}\]\[V+X \frac{dV}{dX}=\frac{2+3V}{3+V}\]\[X \frac{dV}{dX}= \frac{2-V^2}{3+V}\]

Answer 18

u forgot the X when u applied the product rule

Answer 19

\[\frac{3+V}{2-V^2}dV=\frac{dX}{X}\]

Answer 20

oh yes i see now.

Answer 21

:)

Answer 22

x

Answer 23

you found my missing x

Answer 24

lol yay

Answer 25

Awesome i have now arrived at this as my final answer
\[\frac {3}{\sqrt 2}\ln\left|\frac{\sqrt 2 x+y+2-\sqrt 2}{\sqrt2x-y+2-\sqrt 2}\right|=\ln\left|\frac{2x^2-y^2-4x-4y-2}{c}\right|\]

can you think of any reason as why to drop the constant of integration ?

Answer 26

\[\ln|2x^2-y^2-4x-4y-2|-lnC\]
i dont know lol but im thinking

Answer 27

i thought that differential equation always as many constants of integration as the order of the differential equation

Answer 28

i guess c has to be positive

here are some solutions

Answer 29

Here what Mathematica gave as a solution
\[
54 \sqrt{2} \tanh ^{-1}\left(\frac{3 y(x)+2
x+4}{\sqrt{2} (y(x)+3 x-1)}\right)=\\7
c_1+18 \ln \left(\frac{-2 x^2+y(x)^2+4
y(x)+4 x+2}{7 (x-1)^2}\right)+36 \ln
(x-1)
\]
Any professor who ask such a question to be done by hand is likely a masochist ( a person who is gratified by pain, degradation, etc., that is self-imposed or imposed by others).
These days painful problems like this should be done by a machine.
When you try to do it by hand, you write two pages and you forget a sign somewhere and you get a useless solution.

Answer 30

i have found \(\LaTeX \) much easier than by hand,
i have been working through this second-year university text for a few years now,

Answer 31

i failed this differential equations course the first time round

Answer 32

@lalaly you are geniusssssssssss =D

Answer 33

lol wassi im just good at spotting mistakes

Answer 34

lol but you havnt found any mistakes in me till yet :D