Question

Solve the initial value problem \(\large y ' =\dfrac{x^2-y^2 }{xy}\) with y(2)=3 .

Answer 1

Note that the expression on right hand side is of form \(\large f(kx, ky) = k^nf(x, y) \)

Answer 2

try the substitution \(\large \rm y = vx\)

Answer 3

i did but i got the wrong answer

Answer 4

\(\large y = vx \implies y' = v'x + v\)

Answer 5

the equation becomes

\(\large v'x + v = \dfrac{x^2-(vx)^2}{x(vx)}\)


\(\large v'x + v = \dfrac{1-v^2}{v}\)

Answer 6

yes ?

Answer 7

i think when i put the value of y(2)=3 , i made mistake

Answer 8

whats teh general solution you got ?

Answer 9

1x (x 4 +562 ) 12

Answer 10

1/x(X^(4)+56)/2)^(1/2)

Answer 11

is there some change in the question ? whats the original equation ?

Answer 12

\(\large y ' =\dfrac{x^2-y^2 }{xy} \)

right ?

Answer 13

yes

Answer 14

with y(2)= 3

Answer 15

Alright, lets see if i get the same solution as yours :

\(\large \dfrac{dv}{dx}x + v = \dfrac{1-v^2}{v}\)

\(\large \dfrac{dv}{dx}x = \dfrac{1-2v^2}{v} \)

\(\large \int\dfrac{v}{1-2v^2}dv = \int\dfrac{1}{x}dx \)

\(\large -\frac{1}{4}\ln|1-2v^2| = \ln|cx| \)

\(\large (1-2v^2)^{-1/4} = cx\)

\(\large 1-2v^2 = Cx^{-4}\)

\(\large v^2 =\dfrac{1- Cx^{-4}}{2}\)

\(\large v^2 =\dfrac{x^4 - C}{2x^4}\)

Answer 16

plugging in v = y/x back, you get :

\(\large (y/x)^2 = \dfrac{x^4-C}{2x^4}\)

\(\large y^2 = \dfrac{x^4-C}{2x^2}\)

Answer 17

plugin the initial value x = 2, y=3
and solve C

Answer 18

i didnt get why you put v^2 as (x/y)^2 (y/x) 2 =x 4 −C2x 4

Answer 19

second last step

Answer 20

2y^(2)X^(2)=x^(4)-c
c=-x^(2)(2y^2-x^2)

Answer 21

so i got c=-56

Answer 22

so my answer will be y=Square root(x^(4)+56)/(2x^(4))

Answer 23

Determine which of the following pairs of functions are linearly independent.
f(t)=e ^(λt) cos(μt), and g(t)=e ^(λt) sin(μt),μ≠0

Answer 24

that looks good!

Answer 25

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