Question

information about D opprater

Answer 1

D= differential operator
D = d/dx

Answer 2

ok who to solve equation to D operator

Answer 3

you give me an equation

Answer 4

i need information about homogenes

Answer 5

google it simply

Answer 6

ok\[4y+\prime y+ \prime \prime y\]

Answer 7

if its homogeneous must be equal to zero then.
\[\Large y''+y'+4y=0\]

Answer 8

D^2 = y'', D = y'
so u get
(D^2+D+4)y =0
or D^2+D+4=0

Answer 9

by applying D operator
\[\Large (D^2+D+4)y=0\]
you can solve now
\[\Large D^2+D+4=0\]
solve by either quadratic equation or make factors to get its roots.

Answer 10

and one more thing :

\(\huge \color{red}{\text{Welcome to Open Study}}\ddot\smile\)

Answer 11

thnks

Answer 12

i have another equation

Answer 13

ask , we will try our best to help you.

Answer 14

ask it ..

Answer 15

\[^{2} y+^{2} xdiv2xy= \prime y\]

Answer 16

can you plz write it again ?

Answer 17

solve above

Answer 18

and how many solution of this ode pass through( 0,0)

Answer 19

??

Answer 20

result????????
how to solve it??????????????

Answer 21

homogeneous equation
you need to use substitution
\[\Large y=vx\]
using product rule take derivative
\[\Large y'=v+x \frac{dv}{dx}\]
\[\Large y'=\frac{x(vx)}{x^2+(vx)^2}=\frac{x^2v}{x^2+x^2v^2}\]
simplifying
\[\Large y'=\frac{x^2v^2}{x^2(1+v^2)}=\frac{v^2}{1+v^2}\]
put value of y' from above which is \[\Large y'=v+x \frac{dv}{dx}\]
so it becomes
\[\Large v+x \frac{dv}{dx}=\frac{v}{1+v^2}\]
\[\Large x \frac{dv}{dx}=\frac{v}{1+v^2}-v\]
\[\Large x \frac{dv}{dx}=\frac{v-v-v^3}{1+v^2}\]
\[\Large x \frac{dv}{dx}=\frac{-v^3}{1+v^2}\]
which is now separable equation just separate variables
\[\Large \frac{-(1+v^2)}{v^3}dv=\frac{1}{x}dx\]
just integrate both sides
can you now solve this ???

Answer 22

yes

Answer 23

good

Answer 24

ut how many solution of this ode pass through (0,0)

Answer 25

??

Answer 26

ok i have other equation

Answer 27

\[^{2} xexp(-2x)=4y+\prime y4+\prime \prime y\]