Jessie Lee

2021-12-31

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

Anzante2m

Step 1
We have given $Fx\right)=1,{F}^{\prime }\left(0\right)=3,F\left(0\right)=4$
We have to find the function F.
Step 2
We have
$Fx\right)=1$
Take integration on both sides
$\int Fx\right)dx=\int dx$
$⇒{F}^{\prime }\left(x\right)=x+C$
We have given
${F}^{\prime }\left(0\right)=3$
Then, ${F}^{\prime }\left(0\right)=3=0+C$
$⇒C=3$
So, $⇒{F}^{\prime }\left(x\right)=x+3$ (i)
Taking integration on both sides of equation (i) we get,
$\int {F}^{\prime }\left(x\right)dx=\int xdx+\int 3dx$
$⇒F\left(x\right)=\frac{{x}^{2}}{2}+3x+D$
And also we have $F\left(0\right)=4$
Then, $F\left(0\right)=4=\frac{{0}^{2}}{2}+0+D$
$⇒D=4$
Therefore, function is
$F\left(x\right)=\frac{{x}^{2}}{2}+3x+4$

Buck Henry

From the given conditions we have
${F}^{\prime }\left(x\right)=\int 1dx=x+C$
Now if we include that ${F}^{\prime }\left(0\right)=3$, we have
$3={F}^{\prime }\left(0\right)=C$
That is, $C=3$, which means that ${F}^{\prime }\left(x\right)=x+3$
Now we have that $F\left(x\right)=\int \left(x+3\right)dx=\int xdx+\int 3dx=\frac{{x}^{2}}{2}+3x+D$
From the given condition we get
$4=F\left(0\right)=D$
that is, $F\left(x\right)=\frac{{x}^{2}}{2}+3x+4$

karton