Cian Orr

2022-02-18

a) Solve the differential equation:

$(x+1)\frac{dy}{dx}-3y={(x+1)}^{4}$

given that$y=16$ and $x=1$ , expressing the answer in the form of $y=f\left(x\right)$ .

b) Hence find the area enclosed by the graphs$y=f\left(x\right),\text{}y={(1-x)}^{4}$ and the $x-a\xi s$ .

I have found the answer to part a) using the first order linear differential equation method and the answer to part a) is:$y={(1+x)}^{4}$ . However how would you calculate the area between the two graphs ($y={(1-x)}^{4}$ and $y={(1+x)}^{4}$ ) and the x-axis.

given that

b) Hence find the area enclosed by the graphs

I have found the answer to part a) using the first order linear differential equation method and the answer to part a) is:

Jocelyn Harwood

Beginner2022-02-19Added 8 answers

Good job on solving part (a) of the question. That was the tough part. Part (b) is much more easier.

As you stated, we get that the function is:

Now, in order to find the area between this function, the function

When

Next, we look at where each of the two functions intersect the x-axis. We get that:

Therefore, we get our two endpoints: (-1,0) and (1,0).

Thus, we can use construct our integral now:

Hope that helped. Comment if you have any questions.

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My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

$=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}+K$

$K=0$

$A=F(s)\ast (s+a)$