solution of IVP using laplace transform. (D^2 + 2D +1)y = e^(-1) +sinx y(0)=1 y'(0)=0

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

solution of IVP using laplace transform. (D^2 + 2D +1)y = e^(-1) +sinx y(0)=1 y'(0)=0

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

1 Attachment
Is D suppose to be derivative?
no. it's just a variable :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Uh, I'm pretty sure it's the derivative otherwise you have 3 variables which is weird
So all you do is take the laplace of the left hand side and the right hand side going term by term. So the laplace of y'' is Ys^2-y(0)s-y'(0) and the laplace of y'=Ys-y(0) and the laplace of y is just Y. Now for the right hand side, you e^-1 is a constant, and laplace transforms of constants is just 1/C where C is the constant. So the laplace transfrom of e^-1 is just e. You can get the transfrom of sinx from your laplace table. After doing that, simply plug in your values and rearrange to get thigns nice and clean. Then isolate for Y, use partial fractions to split up the giant fraction you get so you can then inverse laplace everything using your table to get back little y.

Not the answer you are looking for?

Search for more explanations.

Ask your own question