# Find the function F that satisfies the following differential equations

Find the function F that satisfies the following differential equations and initial conditions.
$$\displaystyle{F}\text{}{x}{)}={1};{F}'{\left({0}\right)}={3},{F}{\left({0}\right)}={4}$$

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Anzante2m
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
$$\displaystyle{F}{\left({x}\right)}={\frac{{{x}^{{{2}}}}}{{{2}}}}+{3}{x}+{4}$$
###### Not exactly what you’re looking for?
Buck Henry
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, $$\displaystyle{F}{\left({x}\right)}={\frac{{{x}^{{{2}}}}}{{{2}}}}+{3}{x}+{4}$$
karton

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