Cian Orr

2022-02-18

a) Solve the differential equation:
$\left(x+1\right)\frac{dy}{dx}-3y={\left(x+1\right)}^{4}$
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 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: $y={\left(1+x\right)}^{4}$. However how would you calculate the area between the two graphs ($y={\left(1-x\right)}^{4}$ and $y={\left(1+x\right)}^{4}$) and the x-axis.

Jocelyn Harwood

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:
$y={\left(1+x\right)}^{4}$
Now, in order to find the area between this function, the function $f\left(x\right)={\left(1-x\right)}^{4}$, and the x-axis, we need to the find intersection points:
${\left(1+x\right)}^{4}={\left(1-x\right)}^{4}$
${\left({\left(1+x\right)}^{4}\right)}^{\frac{1}{4}}={\left({\left(1-x\right)}^{4}\right)}^{\frac{1}{4}}$
$1+x=1-x$
$x=0$
When $x=0$, ${\left(0+1\right)}^{4}={\left(0-1\right)}^{4}=1$.Thus, we get that the two functions intersect at the point (0,1).
Next, we look at where each of the two functions intersect the x-axis. We get that:
${\left(1+x\right)}^{4}=0$
$1+x=0$
$x=-1$
${\left(1-x\right)}^{4}=0$
$1-x=0$
$x=1$
Therefore, we get our two endpoints: (-1,0) and (1,0).
Thus, we can use construct our integral now:
$A={\int }_{-1}^{0}{\left(1+x\right)}^{4}dx+{\int }_{0}^{1}{\left(1-x\right)}^{4}dx$
${\left[\frac{{\left(x+1\right)}^{5}}{5}\right]}_{\left\{-1\right\}}^{0}+{\left[\frac{{\left(x-1\right)}^{5}}{5}\right]}_{0}^{1}=\frac{1}{5}-0+0-\left(-\frac{1}{5}\right)=\frac{2}{5}$
Hope that helped. Comment if you have any questions.

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