Let T : C∞(R) −→ C∞(R) be the map given by
T(f) = f′′− 2f′− 3f .
Find a linear combination a sin(x) + b cos(x) which solves the equation
f′′− 2f′− 3f = 2 cos(x).
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
you want an exact solution without initial conditions?
or perhaps they just want the particular solution, which you can get with undetermined coefficients
i think so .. this part of the question just asks for a linear combo of sinx and cosx which solves the equation .. then it asks to check if e^-x and e^3x are in the ker, then it says find a sol'n to the same questiona nd gives f(0)=2 and fprime (0) = 3
Not the answer you are looking for? Search for more explanations.
ok, so first you need to apply the method of undetermined coefficients
are you at all familiar with that?
no, my prof gave us this homework but we haven't even looked at it yet.
is that a method in which i'd be able to find if i googled it hah?
no, I'll give you a decent link
basically you are going to guess that there is some solution in the form\[y_p=A\cos x+B\sin x\]and plug that into the equation
the only thing is this is for a lin. alg. course, so i'm not sure if i'm supposed to be using calculus to solve it.
\[y_p'=-A\sin x+B\cos x\]\[y_p''=-A\cos x-B\sin x\]plug that into the original\[y''− 2y'− 3y\]\[ =-A\cos x-B\sin x+2A\sin x-2B\cos x-3A\cos x-3B\sin x\]\[=(-4A-3B)\cos x+(2A-4B)\sin x=2\cos x\]well we are about to get a system from this so hang on
let's compare coefficients of sin and cos on each side
oh ok, that makes sense .. more sense atleast haha
\[-4A-3B=2\]\[2A-4B=0\]now look, it just became linear algebra
so solve for A and B and you will know a linear combination that will solve our problem
I have to go, I hope that helped a bit
the other part, just use the characteristic eqn I guess and you will see that C1e^-x and C2e^3x are solutions
then use the initial conditions about y(0) and y'(0) to find C1 and C2, which will again be linear algebra-ish