Question

For the following differential equations, check that \(y_1\) is a solution of the corresponding homogeneous equation and then calculate the general solution by the reduction of order method.

\[\left(1+x^2\right)y^{\prime\prime}+xy^\prime-y=0;\qquad y_1=x\]

Answer 1

eqaution is not visible

Answer 2

\[\left(1+x^2\right)y^{\prime\prime}+xy^\prime-y=0;\qquad y_1=x\]\[\qquad\qquad\qquad\qquad\qquad\qquad y_1^\prime =1\]\[\qquad\qquad\qquad\qquad\qquad\qquad y_1^{\prime\prime} =0\]
\[\left(1+x^2\right)(0)+x(1)-(x)=0\]
\[\therefore\quad y_1=x \qquad\text{is a solution}\]
\[y_2=uy_1=ux\]\[y_2^\prime=u^\prime x+u\]\[y_2^{\prime\prime}=u^{\prime\prime}x+u^\prime+u^\prime=u^{\prime\prime}x+2u^\prime\]\[\left(1+x^2\right)y^{\prime\prime}_2+xy^\prime_2-y_2=0\]\[\left(1+x^2\right)\left(u^{\prime\prime}x+2u^\prime\right)+x\left(u^\prime x+u\right)-\left(ux\right)=0\]\[\left(1+x^2\right)u^{\prime\prime}x+\left(2+2x^2+x^2\right)u^\prime+\left(x-x\right)u=0\]\[\left(x+x^3\right)u^{\prime\prime}+\left(2+3x^2\right)u^\prime=0\]

Answer 3

am i doing this right , i think i have to solve for \(u\)

Answer 4

\[u(x)\]

Answer 5

you gonnna solve by frobenius?

Answer 6

@UnkleRhaukus
ur doing right solve for u

Answer 7

let u'=p

Answer 8

ah \[\left(x+x^3\right)u^{\prime\prime}+\left(2+3x^2\right)u^\prime=0\] is a \(u\)- absent equation

Answer 9

yes thats right

Answer 10

\[p=u^\prime\]\[p^\prime=u^{\prime\prime}\]

Answer 11

\[\left(x+x^3\right)p^{\prime}+\left(2+3x^2\right)p=0\]\[p^{\prime}+\frac{2+3x^2}{x+x^3}p=0\]

Answer 12

\[\large{\mu(x)=e^{\int{\frac{2+3x^2}{x+x^3}}\text dx}}\]

Answer 13

is this a easy integral ?,

Answer 14

just notice
\[\frac{2+3x^2}{x+x^3}=\frac{1}{x+x^3}+\frac{1+3x^2}{x+x^3}=\frac{1}{x}-\frac{1}{2}\frac{2x}{1+x^2}+\frac{1+3x^2}{x+x^3}\]

Answer 15

you have used partial fractions/

Answer 16

yes it is

Answer 17

ill work on getting that decomposition by myself, ill be back when i have the details

Answer 18

ok

Answer 19

\[\frac{2+3x^2}{x\left(x+x^2\right)}=\frac{1}{x\left(x+x^2\right)}+\frac{1+3x^2}{x\left(x+x^2\right)}\]
\[\frac{1}{x\left(x+x^2\right)}=\frac Ax+\frac B{x(1+x^2)}\]
\[1=x(1+x^2)A+Bx\]

Answer 20

im not how to can find \(A,B\)

Answer 21

* i am not sure how to find \(A,B\)

Answer 22

\[we \ \ have \ \ \frac{2+3x^2}{x+x^3} \ \ \ \ \]
not
\[\frac{2+3x^2}{x(x+x^3)}\]

Answer 23

i reverted back to \[x+x^3=x(1+x^2)\]
to make the partial fractions decomposition easier

Answer 24

i see i have made some mistake s

Answer 25

\[x\left(1+x^2\right)u^{\prime\prime}+\left(2+3x^2\right)u^\prime=0\]\[x\left(1+x^2\right)p^{\prime}+\left(2+3x^2\right)p=0\]\[p^{\prime}+\frac{2+3x^2}{x\left(1+x^2\right)}p=0\]\[\large{\mu(x)=e^{\int{\frac{2+3x^2}{x\left(1+x^2\right)}}\text dx}}\]

Answer 26

\[\frac{2+3x^2}{x\left(1+x^2\right)}=\frac{1}{x\left(1+x^2\right)}+\frac{1+3x^2}{x\left(x+x^2\right)}\]\[\frac{1}{x\left(1+x^2\right)}=\frac Ax+\frac B{x(1+x^2)}\]\[1=x(1+x^2)A+Bx\]

Answer 27

use this
\[\frac{A}{x}+\frac{Bx+C}{1+x^2}\]

Answer 28

ah
clearly my memory is terrible

Answer 29

\[1=x(1+x^2)A+{(Bx+C)}x\]\[1=Ax+Ax^3+Bx^2+Cx\]\[1=Ax^3+Bx^2+(A+C)x\]

Answer 30

how do i find the constants?

Answer 31

i think u made a little mistake in multiply
\[1=(A+B)x^2+Cx+A\]

Answer 32

\[\frac{1}{x\left(1+x^2\right)}=\frac Ax+\frac {Bx+C}{1+x^2}\]\[1=(1+x^2)A+{(Bx+C)}x\]\[1=A+Ax^2+Bx^2+Cx\]\[1=A+(A+B)x^2+Cx\]

Answer 33

so
A=1
?

Answer 34

yes

Answer 35

B=-1 and C=0

Answer 36

how do you find those

Answer 37

A+B=0 and C=0

Answer 38

think of
\[1=0*x^2+0*x+1\]

Answer 39

\[1=A+(A+B)x^2+Cx\]
\[A=1\]\[(A+B)=0;\qquad B=0\]\[C=0\]

Answer 40

**

B=-1

Answer 41

\[\frac{1}{x\left(1+x^2\right)}=\frac Ax+\frac {Bx+C}{1+x^2}=\frac 1x-\frac {x}{1+x^2}\]

Answer 42

exactly

Answer 43

\[\frac{1+3x^2}{x\left(1+x^2\right)}=\frac{1}{x(1+x^2)}+\frac{3x}{1+x^2}\]\[\frac{1}{x(1+x^2)}=\frac Dx+\frac {Ex+F}{1+x^2} \]\[1=D(1+x^2)+(Ex+F)x\]\[1=D+(D+E)x^2+Fx\]
\[D=1\]\[(D+E)=0;\qquad E=-1\]\[F=0\]
\[\frac{1}{x(1+x^2)}=\frac 1x-\frac {x}{1+x^2} \]

Answer 44

\[\int\limits \frac{2+3x^2}{x+x^3}dx=\int\limits (\frac{1}{x}-\frac{1}{2}\frac{2x}{1+x^2}+\frac{1+3x^2}{x+x^3})dx\]

Answer 45

its not Necessary to fractionate
\[\frac{1+3x^2}{x+x^3}\]
because
\[\frac{d}{dx}(x+x^3)=1+3x^2\]

Answer 46

damn your right. that saves a lot of work (That i have already done)
oh well good practice .

Answer 47

\[
(y'(1+x^2))' = (yx)' \\
y'(1+x^2) = yx + c\\
\]

Answer 48

now that's a linear first order DE

Answer 49

thats right but we want to use the reduction of order method

Answer 50

\[\frac{2+3x^2}{x\left(1+x^2\right)}=\frac 1x-\frac {x}{1+x^2}+\frac{1+3x^2}{x\left(1+x^2\right)}\]
\[\frac{2+3x^2}{x\left(1+x^2\right)}=\frac 1x-\frac 12\frac {2x}{1+x^2}+\frac{1+3x^2}{x\left(1+x^2\right)}\]

\[\large{\mu(x)=e^{\int\frac 1x-\frac 12\frac {2x}{1+x^2}+\frac{1+3x^2}{x\left(1+x^2\right)}\text dx}}\]

Answer 51

isn't that the reduction of order??

Answer 52

take a look
http://en.wikipedia.org/wiki/Reduction_of_order

Answer 53

\[\large{\mu(x)=e^{\ln|x|-\ln|1+x^2|+\ln\left|x(1+x^2)\right|}}\]\[=e^{\ln\left|\frac{x^2(1+x^2)}{1+x^2}\right|}\]\[\mu(x)=x^2\]

Answer 54

u forgotted 1/2 check the integral again

Answer 55

your right.

Answer 56

\[\large{\mu(x)=e^{\ln|x|-\ln\sqrt{1+x^2}+\ln\left|x(1+x^2)\right|}}\]
\[=e^{\ln\left|\frac{x^2(1+x^2)}{\sqrt{1+x^2}}\right|}\]
\[\mu(x)=x^2\sqrt{1+x^2}\]

Answer 57

\[\left(x^2\sqrt{1+x^2} p\right)^{\prime}=0\]\[x^2\sqrt{1+x^2}p=1\]\[p=\frac{1}{x^2\sqrt{1+x^2}}\]\[u^{\prime}=\frac{1}{x^2\sqrt{1+x^2}}\]

Answer 58

\[u=\int\frac{1}{x^2\sqrt{1+x^2}}\text dx\]

do i have to use partial fractions decomposition again now///

Answer 59

no

Answer 60

have i made another bunch of mistakes?
or is integration by parts a better idea?
there are integral similar to this in my standard table of [41] integrals
but i cant see any that match

Answer 61

wolfram is suggesting a trig substitution,

Answer 62

sorry i lost my connection about hour ago

Answer 63

yes x=tan t

Answer 64

i dont really like trig substitution because i dont understand why they are valid,

Answer 65

me too

Answer 66

\[u=-\frac{\sqrt{x^2+1}}{x}\]\[y_2=uy_1=-{\sqrt{x^2+1}}\]
\[y=ay_1+by_2\]\[y= ax+b{\sqrt{x^2+1}}\]