Coordinated Differential Equations

We start by taking the $x$ coefficient, $5\text{,}$ and cut it in half and square it, obtaining $\frac{25}{4}\text{.}$ We then add and subtract this term between the $x$ term and the constant, obtaining

The first three terms factor; in this case we take $5$ and cut it in half, and that tells us what to add to $x$ that is squared. We obtain

The $t^2$ coefficient is $4\text{,}$ so we factor it out and get $4(t^2-16t+\frac{5}{2})\text{;}$ we then use the standard completing the square technique on $t^2-16t+\frac{5}{2}\text{.}$ The first step is adding and subtracting $(\frac{-16}{2}{)}^2=64\text{;}$ doing this and factoring gives

Then we multiply this by $4$ to obtain

Rearranging this to equal zero gives $x^2-4x-2=0\text{,}$ and we thus have $a=1,b=-4,c=-2\text{.}$ The solutions are

We have $a=2,b=-12,c=27\text{;}$ plugging this in the quadratic formula gives

We can write

So

Solve the following system of equations using substitution.

We can use either equation to solve for either variable; let’s use the first equation to solve for $y\text{.}$ We get $y=\frac{5-2x}{3}\text{.}$ We then plug this into the second equation to obtain

Simplifying the equation gives $\frac{-16}{3}x+\frac{10}{3}=2\text{;}$ solving this yields $x=\frac{1}{4}\text{;}$ plugging this into $y=\frac{5-2x}{3}$ yields $y=\frac{3}{2}\text{.}$

Solve the following system of equations using elimination.

We want to multiply each equation by something and add them together so that one variable cancels with itself. The $x$ coefficients are $3$ and $2\text{;}$ if we multiply the first equation by $-2$ and the second by $3$ we get a $6x$ and a $-6x$ which will cancel when added.

Doing this, the first equation becomes $-2(3x+5y=2)\to -6x-10y=-4$ and the second equation becomes $3(2x+4y=3)\to 6x+12y=9\text{;}$ adding these equations yields

Solve the system of equations

Using substiting, the first equation tells us that $x=2+2y\text{,}$ and plugging this into the second equation gives $-2(2+2y)+4y=1\text{,}$ which simplifies to $-4=1\text{.}$ This obvious falsehood means there are no solutions.

Solve the system of equations

If we try to eliminate the $x$ variable, we can multiply the first equation by $-3$ and the second by $2\text{,}$ giving $-6x-12y=18$ and $6x+12y=-18\text{;}$ adding these together gives $0=0\text{.}$ This tells us there are infinitely many solutions; to find them, we can take either equation and solve for one of the variables. The first one gives $x=-3-2y\text{,}$ and this is the full set. (That is, to find a solution, you can let $y$ be whatever you want, and then use $x=-3-2y$ to solve for $x\text{.}$)

With this, we compute the derivative and write $du=2e^{x}dx\text{,}$ and solving for $dx$ gives $dx=\frac{du}{2e^x}\text{.}$ Substiting our values of $x$ and $dx$ gives

Substituting back in $u=2e^x+1$ the final answer is

We want $u$ to be something whose derivative appears elsewhere in the expression. It looks like $u=\sin (x)$ and $u=\cos (x)+1$ work in this regard, but it’s best not to have the derivative of $u$ appear in the denominator of a fraction, so $u=\cos (x)+1$ is probably a better choice. We then have $du=-\sin (x)dx$ so $dx=-\frac{du}{\sin (x)}\text{.}$ Therefore,

We substitute $u=\sqrt{x}\text{,}$ which gives $du=\frac{1}{2\sqrt{x}}dx$ so $dx=2\sqrt{x}du\text{.}$ Also, the bounds of the integral go from $x=\frac{{\pi }^2}{4}$ to $x={\pi }^2\text{;}$ the substitution means the integral now goes from $u=\frac{\pi }{2}$ to $u=\pi \text{.}$ We therefore get

We do $u=x+1,v^{\prime }=e^x\text{,}$ which gives $u^{\prime }=1,v=e^x\text{;}$ the formula therefore says

We do $u=\sin (x),v^{\prime }=e^{2x}\text{,}$ which gives $u^{\prime }=\cos (x),v=\frac{e^{2x}}{2}\text{,}$ so

We now have to calculate $\int e^{2x}\cos (x)dx\text{.}$ Using integration by parts again, we have $u=\cos (x),v^{\prime }=e^{2x}\text{,}$ so $u^{\prime }=-\sin (x),v=\frac{e^{2x}}{2}\text{,}$ and we get

It looks like we are back at square one, but this is actually helpful. We can write

If we denote $I=\int e^{2x}\sin (x)dx\text{,}$ this equation can be written as

solving this equation for $I$ yields

We do $u=\ln (x),v^{\prime }=x^2\text{,}$ which gives us $u^{\prime }=\frac{1}{x},v=\frac{x^3}{3}\text{,}$ so this ends up being

The denominator factors as $(x-3)(x+1)\text{,}$ so we want to write this as $\frac{8x-20}{x^2-2x-3}=\frac{A}{x-3}+\frac{B}{x+1}\text{;}$ multiplying both sides by $(x-3)(x+1)$ gives

There are two ways to solve this; we can expand out the right-hand-side and equate like terms, or pick "smart" choices for $x$ that help us solve. Both methods are useful in different situations, so I will show both.

Instead of multiplying out, we could have noted that plugging in $x=-1$ into the equation eliminates the $A$ variable, and we get $-28=-4B\text{,}$ so $B=7\text{,}$ and plugging in $x=3$ gives $4=4A\text{,}$ so $A=1\text{.}$ We then have

Factoring the denominator gives $x^3+2x^2+x=x(x^2+2x+1)=x(x+1{)}^2\text{.}$ The partial fractions split-up will therefore be of the form $\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1{)}^2}\text{.}$ Then we take the equation $\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1{)}^2}=\frac{3x^2+3x+1}{x^3+2x^2+x}$ and multiply both sides by $x(x+1{)}^2\text{,}$ which yields

Plugging in $x=0$ gives $A=1\text{.}$ Plugging in $x=-1$ gives $-C=1\text{,}$ so $C=-1\text{.}$ Then plugging in $x=1$ gives $4A+2B+C=7\text{,}$ So $3+2B=7\text{,}$ so $B=2\text{.}$ Therefore,

The best we can factor the denominator is $x^3-2x^2+5x$$=x(x^2-2x+5)\text{.}$ As $x^2-2x+5$ does not factor, our backup plan is to complete the square with it, giving $(x-1{)}^2+4\text{.}$ The partial fraction split-up is therefore

Multiplying both sides by $x((x-1{)}^2+4)$ gives

Plugging in $x=0$ gives $15=5A\text{,}$ so $A=3\text{.}$ Plugging in $x=1$ gives $4A+C=14\text{;}$ since $A=3$ this yields $C=2\text{.}$ Finally, plugging in $x=2$ gives $21=5A+2B+2C\text{,}$ so $B=1\text{.}$ Therefore,

To evaluate that last integral, we split it into

The first term can be integrated by substituting $u=(x-1{)}^2+4\text{,}$ and we get $\frac{1}{2}\ln ((x-1{)}^2+4)\text{.}$ For the second term, recall that $\int \frac{1}{t^2+1}dt=\arctan (t)+C\text{.}$ Since $4=2^2$ we can substitute $x-1=2u$ and find the second term integrates to $\arctan (\frac{x-1}{2})\text{,}$ so the final answer is