Question

Find solutions y1 and y2 of the equation y"=0 that satisfy the initial conditions y1(xo)=1, y'1(xo)=0 and y2(xo)=0, y'2(xo)=1.

Answer 1

@ganeshie8 @ParthKohli @iambatman @zepdrix

Answer 2

@nincompoop @pooja195 @Directrix @Nnesha

Answer 3

Wait, you said to integrate y"=0, right?

Answer 4

Ya that would be a good first step :)
To integrate it twice.

I'm still thinking about this one though XD

Answer 5

It looks like you asked this question 6 months ago lol
I'm looking at it currently, seeing what was discussed.

Answer 6

\[y'=c\]

Answer 7

\[y=c _{1}+c _{2}x\]

Answer 8

Now what? What do I do?

Answer 9

K cool.
I guess we could say this: \(\large\rm y=y_1+y_2\)
So we can write our equation like this: \(\large\rm y_1+y_2=c_1+c_2 x\)
Differentiating gives us: \(\large\rm y_1'+y_2'=c_2\)

Answer 10

And then use the initial data to figure out some stuff.

Answer 11

Given that, \(\large\rm \color{royalblue}{y_1(x_o)=1}\) and \(\large\rm \color{orangered}{y_2(x_o)=0}\)
We have:\[\large\rm \color{royalblue}{y_1(x_o)}+\color{orangered}{y_2(x_o)}=\color{royalblue}{1}+\color{orangered}{0}=c_1+c_2 x_o\]

Answer 12

So that gives us an equation for our constants,\[\large\rm 1=c_1+c_2 x_o\]

Answer 13

Also given, \(\large\rm \color{royalblue}{y_1'(x_o)=0}\) and \(\large\rm \color{orangered}{y_2'(x_o)=1}\)
We have:\[\large\rm \color{royalblue}{y_1'(x_o)}+\color{orangered}{y_2'(x_o)}=\color{royalblue}{0}+\color{orangered}{1}=c_2\]Which gives us\[\large\rm c_2=1\]

Answer 14

Plug this information back into the other constant equation,\[\large\rm 1=c_1+c_2 x_o\qquad\to\qquad 1=c_1+1x_o\]Solving for c_1 gives us,\[\large\rm c_1=1-x_o\]And recall that we also have:\[\large\rm c_2=1\]

Answer 15

If we go all the way back to our equation:\[\large\rm y_1+y_2=c_1+c_2 x\]we have:\[\large\rm y_1+y_2=(1-x_o)+1 x\]

Answer 16

And it looks like the last time you asked this question,
you had said that the solutions are supposed to be:
\(\large\rm y_1=1,\qquad y_2=x-x_0\)

Is that still the case? Do you have access to the answer?
Seems like the solutions should be:
\(\large\rm y_1=1-x_o,\qquad y_2=x\)

Hmm..

Answer 17

Yes, well, the book has the answers y1=1, y2=x-x0. But are you really sure that your answers are correct? The book answers are wrong?

Answer 18

No, I'm not confident enough to say that the book is wrong,
I'm a little rusty on my ODE's. :)

I could be wrong, seems more likely that the book answer is incorrect though.
That's what Sith had suggested last time as well it seems.

Answer 19

Dude, I finally found the way to do this problem and the book's answers are right! Thank you for the help though. :)

Answer 20

Oh interesting :OO
Now I'm kind of curious lol