Find the function F that satisfies the following differential equations

abreviatsjw 2021-12-31 Answered
Find the function F that satisfies the following differential equations and initial conditions.
\(\displaystyle{F}{''}{\left({x}\right)}={\cos{{x}}},{F}'{\left({0}\right)}={3},{F}{\left(\pi\right)}={5}\)

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Expert Answer

Cassandra Ramirez
Answered 2022-01-01 Author has 136 answers
Step 1
Given that,
(1) \(\displaystyle{f}{''}{\left({x}\right)}={\cos{{x}}}\)
(2) PKf'(0)=3ZSK
(3) \(\displaystyle{f{{\left(\pi\right)}}}={5}\)
Integrate equation 1;
\(\displaystyle\int{f}\text{}{x}{)}{\left.{d}{x}\right.}=\int{\cos{{\left({x}\right)}}}{\left.{d}{x}\right.}\)
\(\displaystyle{f}'{\left({x}\right)}={\sin{{\left({x}\right)}}}={\sin{{\left({x}\right)}}}+{c}\) (4)
Using criteria (2);
\(\displaystyle{f}'{\left({0}\right)}={\sin{{\left({0}\right)}}}+{c}\)
\(\displaystyle{3}={0}+{c}\)
\(\displaystyle\Rightarrow{c}={3}\)
Equation (4) is are written as,
\(\displaystyle{f}'{\left({x}\right)}={\sin{{\left({x}\right)}}}+{3}\) ..(5)
Step 2
Integrate equation 5;
\(\displaystyle\int{f}'{\left({x}\right)}{\left.{d}{x}\right.}=\int{\left[{\sin{{\left({x}\right)}}}+{3}\right]}{\left.{d}{x}\right.}\)
\(\displaystyle{f{{\left({x}\right)}}}=-{\cos{{\left({x}\right)}}}+{3}{x}+{c}'\) (5)
Using criteria (3),
\(\displaystyle{f{{\left(\pi\right)}}}=-{\cos{{\left(\pi\right)}}}+{3}\pi+{c}'\)
\(\displaystyle{5}={1}+{3}\pi+{c}'\)
\(\displaystyle\Rightarrow{c}'={4}-{3}\pi\)
Equation (5) is are written as,
\(\displaystyle{f{{\left({x}\right)}}}=-{\cos{{\left({x}\right)}}}+{3}{x}+{4}-{3}\pi\)
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Nadine Salcido
Answered 2022-01-02 Author has 1327 answers
As we know \(\displaystyle{F}\text{}{x}{)}={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{F}'{\left({x}\right)}\). Therefore, F'(x) is an antiderivative of F"(x).
\(\displaystyle{F}'{\left({x}\right)}=\int{F}\text{}{x}{)}{\left.{d}{x}\right.}=\int{\cos{{x}}}{\left.{d}{x}\right.}={\sin{{x}}}+{C}_{{{1}}}\) (2)
Taking into account that \(\displaystyle{F}'{\left({0}\right)}={3}\) we have:
\(\displaystyle{F}'{\left({0}\right)}={3}\Rightarrow{\sin{{\left({0}\right)}}}+{C}_{{{1}}}={3}\Rightarrow{C}_{{{1}}}={3}\) (3)
Then \(\displaystyle{F}'{\left({x}\right)}={\sin{{x}}}+{3}\)
Since \(\displaystyle{F}'{\left({x}\right)}={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{F}{\left({x}\right)}\) we have that:
\(\displaystyle{F}{\left({x}\right)}=\int{F}'{\left({x}\right)}{\left.{d}{x}\right.}=\int{\left({\sin{{x}}}+{3}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle=\int{\sin{{x}}}{\left.{d}{x}\right.}+\int{3}{\left.{d}{x}\right.}\) (4)
\(\displaystyle=-{\cos{{x}}}+{3}{x}+{C}_{{{2}}}\)
Since \(\displaystyle{F}{\left(\pi\right)}={4}\) we have that
\(\displaystyle{F}{\left(\pi\right)}={4}\Rightarrow-{\cos{{\left(\pi\right)}}}+{3}{\left(\pi\right)}+{C}_{{{2}}}={4}\)
\(\displaystyle\Rightarrow{1}+{3}\pi+{C}_{{{2}}}={4}\) (5)
\(\displaystyle\Rightarrow{C}_{{{2}}}={3}{\left({1}-\pi\right)}\)
Finally, we have found that
\(\displaystyle{F}{\left({x}\right)}=-{\cos{{x}}}+{3}{x}+{3}{\left({1}-\pi\right)}\) (6)
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karton
Answered 2022-01-10 Author has 8454 answers

From the given conditions we have
\(F'(x)=\int \cos xdx=\sin x+C\)
Now if we include that \(F'(0)=3\), we have
\(3=F'(0)=C\)
that is, \(C=3\), which means that \(F'(x)=\sin x+3\)
Now we have that
\(F(x)=\int (\sin x+3)dx=\int \sin xdx+\int 3dx=-\cos x+3x+D\)
From the given condition we get
\(4=F(\pi)=1+3 \pi+D\)
that is, \(D=3-3 \pi\), so \(F(x)=-\cos x+3x+3-3 \pi\)

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