Jessie Lee

2021-12-31

Find the function F that satisfies the following differential equations and initial conditions.

$Fx)=1;{F}^{\prime}\left(0\right)=3,F\left(0\right)=4$

Anzante2m

Beginner2022-01-01Added 34 answers

Step 1

We have given$Fx)=1,{F}^{\prime}\left(0\right)=3,F\left(0\right)=4$

We have to find the function F.

Step 2

We have

$Fx)=1$

Take integration on both sides

$\int Fx)dx=\int dx$

$\Rightarrow {F}^{\prime}\left(x\right)=x+C$

We have given

${F}^{\prime}\left(0\right)=3$

Then,${F}^{\prime}\left(0\right)=3=0+C$

$\Rightarrow C=3$

So,$\Rightarrow {F}^{\prime}\left(x\right)=x+3$ (i)

Taking integration on both sides of equation (i) we get,

$\int {F}^{\prime}\left(x\right)dx=\int xdx+\int 3dx$

$\Rightarrow F\left(x\right)=\frac{{x}^{2}}{2}+3x+D$

And also we have$F\left(0\right)=4$

Then,$F\left(0\right)=4=\frac{{0}^{2}}{2}+0+D$

$\Rightarrow D=4$

Therefore, function is

$F\left(x\right)=\frac{{x}^{2}}{2}+3x+4$

We have given

We have to find the function F.

Step 2

We have

Take integration on both sides

We have given

Then,

So,

Taking integration on both sides of equation (i) we get,

And also we have

Then,

Therefore, function is

Buck Henry

Beginner2022-01-02Added 33 answers

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$F\left(x\right)=\int (x+3)dx=\int xdx+\int 3dx=\frac{{x}^{2}}{2}+3x+D$

From the given condition we get

$4=F\left(0\right)=D$

that is,$F\left(x\right)=\frac{{x}^{2}}{2}+3x+4$

Now if we include that

That is,

Now we have that

From the given condition we get

that is,

karton

Expert2022-01-10Added 613 answers

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