anonymous
  • anonymous
Can anyone help me to solve this differential equation: (-d/dx)(15*dT(X)/dX)+15*T(x)=528.75
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
The LHS is:\[-\frac{d}{dx}\left( 15\frac{dT}{dx} \right)+15T=-15\frac{d^2T}{dx^2}+15T\]so, letting 528.75=c, and dividing both sides by -15, you have,\[T''-T=-\frac{c}{15} \rightarrow T''-\left( T-\frac{c}{15} \right)=0\]Let\[v=T-\frac{c}{15} \]then\[v''=T''\]and you have\[v''-v=0\]which is first order, homogeneous, with constant coefficients, so you can assume a solution of the form\[v=e^{\lambda x}\]Hence\[v''=\lambda^2 e^{\lambda x}\]and substituting into the d.e., we have\[e^{\lambda x}(\lambda^2-1)=0 \rightarrow \lambda = \pm 1\]So the solution is\[v=c_1e^x+c_2e^{-x}\]But \[v=T-\frac{c}{15}\]so\[T=c_1e^x+c_2e^{-x}+\frac{c}{15}\]where c=528.75.
anonymous
  • anonymous
thank you very much...
anonymous
  • anonymous
You're welcome.

Looking for something else?

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

More answers

anonymous
  • anonymous
-d/dx(15*(dT/dx))-d/dy(15*(dt/dy))=30 about this D.E. i can do the same steps like for the previews one?

Looking for something else?

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