# Find the function F that satisfies the following differential equations

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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Cassandra Ramirez
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,
$$\displaystyle{f{{\left({x}\right)}}}=-{\cos{{\left({x}\right)}}}+{3}{x}+{4}-{3}\pi$$
###### Not exactly what youâ€™re looking for?
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)
karton

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 $$F(x)=-\cos x+3x+3-3 \pi$$