Second order differential equation

[Garlick25]

could someone help solve this for me please. Been trying for a week now and not understanding the methods. Thanks in advance

 

[Zermelo]

In this equation, you have y’ and y’’, and no y. That means that you can look at it as a first order differential equation of the function y’. Or simply, make the substitition y’ = z, see what comes up, and get back to us.

 

[Deleted member 4993]

View attachment 32161

could someone help solve this for me please. Been trying for a week now and not understanding the methods. Thanks in advance

First of all you need to decide whether you have a initial value problem or boundary value problem? What methods have you been taught for such problems. Then read the response #2 carefully and tell us what you found?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem.

 

[Garlick25]

This is the whole problem and what i have so far. Thank

 

[Deleted member 4993]

This is the whole problem and what i have so far. ThankView attachment 32162
View attachment 32165

C₁ - t² = ( √C₁ + t) * (√C₁ - t)

use partial fractions and continue......

 

[BigBeachBanana]

Or factor out [imath]c_1[/imath].
[math]y=\int - \frac{1}{-t^2+c_1}\, dt =\int \frac{1}{t^2+c_1}\,dt= \int \frac{1}{c_1}\cdot \frac{1}{\frac{t^2}{c_1}+1}\, dt[/math]Now, let [imath]u=\frac{t}{\sqrt{c_1}}[/imath]. After the substitution, you might be able to recognize that integral.