Find the function F that satisfies the following differential equations

Jessie Lee 2021-12-31 Answered
Find the function F that satisfies the following differential equations and initial conditions.
\(\displaystyle{F}\text{}{x}{)}={1};{F}'{\left({0}\right)}={3},{F}{\left({0}\right)}={4}\)

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Expert Answer

Anzante2m
Answered 2022-01-01 Author has 256 answers
Step 1
We have given \(\displaystyle{F}\text{}{x}{)}={1},{F}'{\left({0}\right)}={3},{F}{\left({0}\right)}={4}\)
We have to find the function F.
Step 2
We have
\(\displaystyle{F}\text{}{x}{)}={1}\)
Take integration on both sides
\(\displaystyle\int{F}\text{}{x}{)}{\left.{d}{x}\right.}=\int{\left.{d}{x}\right.}\)
\(\displaystyle\Rightarrow{F}'{\left({x}\right)}={x}+{C}\)
We have given
\(\displaystyle{F}'{\left({0}\right)}={3}\)
Then, \(\displaystyle{F}'{\left({0}\right)}={3}={0}+{C}\)
\(\displaystyle\Rightarrow{C}={3}\)
So, \(\displaystyle\Rightarrow{F}'{\left({x}\right)}={x}+{3}\) (i)
Taking integration on both sides of equation (i) we get,
\(\displaystyle\int{F}'{\left({x}\right)}{\left.{d}{x}\right.}=\int{x}{\left.{d}{x}\right.}+\int{3}{\left.{d}{x}\right.}\)
\(\displaystyle\Rightarrow{F}{\left({x}\right)}={\frac{{{x}^{{{2}}}}}{{{2}}}}+{3}{x}+{D}\)
And also we have \(\displaystyle{F}{\left({0}\right)}={4}\)
Then, \(\displaystyle{F}{\left({0}\right)}={4}={\frac{{{0}^{{{2}}}}}{{{2}}}}+{0}+{D}\)
\(\displaystyle\Rightarrow{D}={4}\)
Therefore, function is
\(\displaystyle{F}{\left({x}\right)}={\frac{{{x}^{{{2}}}}}{{{2}}}}+{3}{x}+{4}\)
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Buck Henry
Answered 2022-01-02 Author has 4502 answers
From the given conditions we have
\(\displaystyle{F}'{\left({x}\right)}=\int{1}{\left.{d}{x}\right.}={x}+{C}\)
Now if we include that \(\displaystyle{F}'{\left({0}\right)}={3}\), we have
\(\displaystyle{3}={F}'{\left({0}\right)}={C}\)
That is, \(\displaystyle{C}={3}\), which means that \(\displaystyle{F}'{\left({x}\right)}={x}+{3}\)
Now we have that \(\displaystyle{F}{\left({x}\right)}=\int{\left({x}+{3}\right)}{\left.{d}{x}\right.}=\int{x}{\left.{d}{x}\right.}+\int{3}{\left.{d}{x}\right.}={\frac{{{x}^{{{2}}}}}{{{2}}}}+{3}{x}+{D}\)
From the given condition we get
\(\displaystyle{4}={F}{\left({0}\right)}={D}\)
that is, \(\displaystyle{F}{\left({x}\right)}={\frac{{{x}^{{{2}}}}}{{{2}}}}+{3}{x}+{4}\)
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karton
Answered 2022-01-10 Author has 8454 answers

\(\begin{array}{} F'(x)=\int 1dx=x+C \\F'(0)=3 \\3=F'(0)=C \\C=3,\ \text{which means that}\ F'(x)=x+3 \\F(x)=\int (x+3)dx=\int xdx+\int 3dx=\frac{x^{2}}{2}+3x+D \\4=F(0)=D \\F(x)=\frac{x^{2}}{2}+3x+4 \end{array}\)

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