Question

Linear equation of order one..

Answer 1

\[(1+t^2)ds + 2t(st^2 -3(1+t^2)^2)dt=0\]

Answer 2

\[\frac{ ds }{ dt }+ \frac{ 2t^3 }{ 1+t^2 }(s)=3(1+t^2)\]

Answer 3

I.F. =\[\int\limits_{}^{}\frac{ 2t^3 }{ 1+t^2 }dt +k\]

Answer 4

\[\frac{ds}{dt}+\frac{2t^3}{1+t^2}=6(1+t^2)\]

Answer 5

how can i integrate this, lol im confused

Answer 6

oh yeah i forgot, that one

Answer 7

Integrating factor \[e^{\int \frac{2t^3}{1+t^2} dt}\]

Answer 8

can i separate this one and do like this? \[\int\limits_{}^{}\frac{ t^3 }{ 1 }+\frac{ t^3 }{ t^2}\]

Answer 9

Nope

Answer 10

quotient then?

Answer 11

lol , i dont know even the quotient

Answer 12

\[2\int \frac{t^2}{t^2+1} tdt\]

Answer 13

Now substitute t^2+1=u then 2tdt=du

Answer 14

\[\int \frac{u-1}{u} du\]

Answer 15

how did you get that int. u-1/u du?

Answer 16

u+ lnu-1

Answer 17

\[(t^2 +1)+\ln (t^2) +k\]

Answer 18

\[e ^{(t^2+1)}+t^2\]

Answer 19

\[s(t^2e ^{t^2+1})=\int\limits_{}^{}3(1+t^2)(t^2e ^{t^2+1})\]

Answer 20

can anyone help me integrate the RHS

Answer 21

@jiteshmeghwal9

Answer 22

this is my last problem, can anyone help

Answer 23

\[\int \frac{t^2}{t^2+1} 2tdt\] t^2+1= u
=>2tdt=dy
=>t^2=u-1 \[\int \frac{u-1}{u} du\]

Answer 24

= u-ln u + k
Replacing u by t^2+1
= \[(t^2+1)-ln(t^2+1)+k\]

Answer 25

yup i got \[e ^{(t^2+1)+\ln(t^2)}\]

Answer 26

\[\text{I.F.}=e^{\frac{2t^3}{t^2+1}}=e^{(t^2+1)-ln(t^2+1)+k}\]

Answer 27

Finally \[\text{I.F.}=\frac{e^{t^2+1}}{t^2+1}\]

Answer 28

Solution form \[y* I.F. = \int Q.* I.F. dt +k\]

Answer 29

\[y* \frac{e^{t^2+1}}{t^2+1}= \int 6(t^2+1)* \frac{e^{t^2+1}}{t^2+1}\]

Answer 30

Got it up to this step ?

Answer 31

yis.. ^.^

Answer 32

how can i integrate this \[\int\limits_{}^{}te ^{t^2+1}\]

Answer 33

i got \[3e ^{t^2+1} +c\]

Answer 34

I think the RHS should be 6t(1+t^2)

Answer 35

jitesh - correct me if i'm wrong but doesn't the IF = e^t^2 / ( 1 + t^2)

Answer 36

Yeah u r right @welshfella

Answer 37

U figured out my mistake

Answer 38

The ODE should be
\[\dfrac{ds}{dt}+\dfrac{2t^3}{1+t^1}s =6t(1+t^2)\]

Answer 39

yes - I missed that

Answer 40

gtg

Answer 41

the answer should be \[s= (1+t^2)(3-\exp(-t^2))\]

Answer 42

i dont get it ~.~ and its 2:00 am here....BTW this is my assignment for tommorow...

Answer 43

how is the I.F. ? i dont get it lol

Answer 44

you didnt times the I.F. on RHS

Answer 45

\[y*\frac{e^{t^2}}{t^2+1}= \int 6t(1+t^2)* \frac{e^{t^2}}{t^2+1} +k \]

Answer 46

\[y*\frac{e^{t^2}}{t^2+1}=\int 6t e^{t^2} dt + k \]

Answer 47

RHS \[\int\limits_{}^{}6te ^{t^2}+k\]

Answer 48

Solving R.H.S
Substituting t^2=u then
2tdt=du
\[\int 3 e^u du \]
\[3e^u \]
\[3e^{t^2}\]

Answer 49

when t=0, s=2...... i forgot this

Answer 50

how did you get the I.F. = thats my only question now ^.^

Answer 51

@sammixboo

Answer 52

can anyone can integrate \[e ^{\int\limits_{}^{}}\frac{ 2t^3 }{ 1+t^2 }dt\]

Answer 53

they get,how they get it? \[\frac{ e ^{t^2} }{ t^2+1 }\]

Answer 54

INT 2t^3 t^3
----- dt = 2 INT -------
1 + t^2 1 + t^2

Now divide the fraction
= 2 INT ( t - t / (t^2 + 1)
= 2 * t^2 /2 - 1/2 ln (t^2 + 1)

= t^2 - ln (t^2 + 1)

Answer 55

Taking this to the power e:-

= e^ [(t^2 - ln (t^2 + 1)]

= e^(t^2 0
-----
e^ln(t^2 + 1) (Now this equal t^2 + 1)
so we have

e^t^2
-------
t^2 + 1

Answer 56

* the zero in the third line should be a )