Find the particular solution that satisfies the differential equation and the in

Marvin Mccormick 2021-10-29 Answered
Find the particular solution that satisfies the differential equation and the initial condition. \(\displaystyle{f}”{\left({x}\right)}={\sin{{x}}},{f}’{\left({0}\right)}={1},{f{{\left({0}\right)}}}={6}\)

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question

Expert Answer

escumantsu
Answered 2021-10-30 Author has 8934 answers
Step 1
\(\displaystyle\text{We have: }\ {f}{''}{\left({x}\right)}={\sin{{x}}},{f}'{\left({0}\right)}={1},{f{{\left({0}\right)}}}={6}\)
\(\displaystyle\text{We integrate }\ {f}{''}{\left({x}\right)}={\sin{{x}}}\ \text{ (first) and }\ {f}'{\left({0}\right)}\ \text{ (second) to obtain a generalized solution:}\)
\(\displaystyle\int{f}{''}{\left({x}\right)}{\left.{d}{x}\right.}=\int{\sin{{x}}}{\left.{d}{x}\right.}=-{\cos{{x}}}+{C}_{{1}}\)
Step 2
Now, we are applying the initial condition \(\displaystyle{f}'{\left({0}\right)}={1}:\)
\(\displaystyle-{\cos{{0}}}+{C}_{{1}}={1}\)
\(\displaystyle-{1}+{C}_{{1}}={1}\)
\(\displaystyle{C}_{{1}}={1}+{1}\)
\(\displaystyle{C}_{{1}}={2}\)
\(\displaystyle\text{So: }\ {f}'{\left({x}\right)}=-{\cos{{x}}}+{2}\)
Now, we integrate f`(x):
\(\displaystyle\int{f}'{\left({x}\right)}{\left.{d}{x}\right.}=\int{\left(-{\cos{{x}}}+{2}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle=\int-{\cos{{x}}}{\left.{d}{x}\right.}+\int{2}{\left.{d}{x}\right.}\)
\(\displaystyle=-{\sin{{x}}}+{2}{x}+{C}_{{2}}\)
Now, we are applying the initial condition f(0)=6:
\(\displaystyle{\sin{{0}}}+{2}\cdot{0}+{C}_{{2}}={6}\)
\(\displaystyle{0}+{C}_{{2}}={6}\)
\(\displaystyle{C}_{{2}}={6}\)
\(\displaystyle\text{So: }\ {f{{\left({x}\right)}}}=-{\sin{{x}}}+{2}{x}+{6}\)
Finaly, the particular solution is:
\(\displaystyle{f{{\left({x}\right)}}}=-{\sin{{x}}}+{2}{x}+{6}\)
Step 3
Explanetion:
\(\displaystyle{\left({1}\right)}:{f}{''}{\left({x}\right)}={\sin{{x}}}\)
\(\displaystyle{\left({2}\right)}:\int{\sin{{x}}}{\left.{d}{x}\right.}=-{\cos{{x}}}+{C}\)
\(\displaystyle{\left({3}\right)}:{f}'{\left({x}\right)}=-{\cos{{x}}}+{2}\)
\(\displaystyle{\left({4}\right)}:\int{\cos{{x}}}{\left.{d}{x}\right.}={\sin{{x}}}+{C}\)
Result
\(\displaystyle{f{{\left({x}\right)}}}=-{\sin{{x}}}+{2}{x}+{6}\)
Have a similar question?
Ask An Expert
0
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-10-18

Find the solution of the differential equation that satisfies the given initial condition. \(dy/dx=x/y, y(0)=-3\)

asked 2021-03-21
Find the general solution for each differential equation. Verify that each solution satisfies the original differential equation.
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={4}{e}^{{-{3}{x}}}\)
asked 2021-10-28
Show that the equation has exactly one real root.
\(\displaystyle{2}{x}+{\cos{{x}}}={0}\)
asked 2021-09-16

Find a suitable form for particular solution of the differential equation
\(\displaystyle{y}′′+{4}{y}={e}−{2}{t}{\left({t}^{2}+{1}\right)}\)

asked 2021-05-21
Prove that \(\displaystyle{1}-{\left(\frac{{{\cos}^{{2}}{\left({x}\right)}}}{{1}}+{\sin{{\left({x}\right)}}}\right)}\)
asked 2021-05-28
Given that \(\displaystyle{f{{\left({x}\right)}}}={\cos{{x}}}\), show that \(\displaystyle\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}={\cos{{x}}}{\left(\frac{{{\text{cosh}{-}}{1}}}{{h}}\right)}+{\sin{{x}}}{\left(\frac{{\text{sinh}}}{{h}}\right)}\)
asked 2021-05-02
Given that \(\displaystyle{t}{\left({x}\right)}={\sin{{x}}}\), show that \(\displaystyle{f{{\left({x}+{h}\right)}}}-\frac{{f{{\left({x}\right)}}}}{{h}}={\sin{{x}}}{\left(\frac{{{\text{cosh}{-}}{1}}}{{h}}\right)}+{\cos{{x}}}{\left(\frac{{\text{sinh}}}{{h}}\right)}\)

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question
...