anonymous
  • anonymous
d^2y/dt^2= 1-e^2t, y(1)=-1,y'(1)=0 What is the initial value?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
myininaya
  • myininaya
Do you mean solve the IVP?
anonymous
  • anonymous
Yes:)
anonymous
  • anonymous
The way I approach such a problem: Solve the homogenous part of the differential equation first: \[\Large y''(x)=0 \] So the solution to this, almost trivial DE is: \[\Large y_g(x)=Cx+K \] Where C and K are constants of integration. Now you need to find the particular solution to the homogenous differential equation. This is usually done by guessing. Since: \[\Large y''(x)=1-e^{2x} \] A 'okay' guess could be: \[\Large y_p(x)=A+Be^{2x} \] Notice here that this is not yet good enough, see the solution already obtained above. We have Cx + K. So, in order to not overlap any solutions, we want to increase the linear exponent again: \[\Large y_p(x)=Ax^2+Be^{2x} \] Now this is a good guess. Substituting back will give you: \[\Large 2A+4Be^{2x}=1-e^{2x} \] From which you can obtain that: \[ \Large A= \frac{1}{2}, \ B= -\frac{1}{4}\] So given by the principle of superposition you have: \[\Large y(x)=y_p(x)+y_g(x)=Cx+K+\frac{1}{2}x^2-\frac{1}{4}e^{2x} \]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
I am not yet familiar with the methods you demonstrated above. Here is the answer I get: \[^{x^2/2}-^{e^2x/4}+^{e^2/4}-x-(1/2)\]
anonymous
  • anonymous
As I am new to typing equations, I wonder how you made your's look so neat?
anonymous
  • anonymous
I recommend you to check out Paul's website then, he has an amazing introduction to the method I suggested above, clearly, it's not the only one to solve this problem. But it makes inhomogeneous differential equations quite straight forward to solve. Here is the link: http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx As to \(\LaTeX\). First I use the perimeter \Large to make my equations look bigger, or in the case of this website - just more readable. You can write fraction by typing \frac{nominator}{denominator}
anonymous
  • anonymous
Thanks for the help:)

Looking for something else?

Not the answer you are looking for? Search for more explanations.