These equations are nonlinear because of terms like
${\left({y}^{\prime}\right)}^{4},{y}^{3},$ etc. Due to these terms, it is impossible to put these equations into the same form as
[link] .
Our main goal in this section is to derive a solution method for equations of this form. It is useful to have the coefficient of
${y}^{\prime}$ be equal to
$1.$ To make this happen, we divide both sides by
$3{x}^{2}-4.$
This is called the
standard form of the differential equation. We will use it later when finding the solution to a general first-order linear differential equation. Returning to
[link] , we can divide both sides of the equation by
$a\left(x\right).$ This leads to the equation
Now define
$p\left(x\right)=\frac{b\left(x\right)}{a\left(x\right)}$ and
$q\left(x\right)=\frac{c\left(x\right)}{a\left(x\right)}.$ Then
[link] becomes
${y}^{\prime}+p(x)y=q(x).$
We can write any first-order linear differential equation in this form, and this is referred to as the standard form for a first-order linear differential equation.
Writing first-order linear equations in standard form
Put each of the following first-order linear differential equations into standard form. Identify
$p\left(x\right)$ and
$q\left(x\right)$ for each equation.
$y\prime =3x-4y$
$\frac{3xy\prime}{4y-3}=2$ (here
$x>0)$
$y=3y\prime -4{x}^{2}+5$
Add
$4y$ to both sides:
$y\prime +4y=3x.$
In this equation,
$p\left(x\right)=4$ and
$q\left(x\right)=3x.$
Multiply both sides by
$4y-3,$ then subtract
$8y$ from each side:
Finally, divide both sides by
$3x$ to make the coefficient of
$y\prime $ equal to
$1\text{:}$
$y\prime -\frac{8}{3x}y=-\frac{2}{3x}.$
This is allowable because in the original statement of this problem we assumed that
$x>0.$ (If
$x=0$ then the original equation becomes
$0=2,$ which is clearly a false statement.)
In this equation,
$p\left(x\right)=-\frac{8}{3x}$ and
$q\left(x\right)=-\frac{2}{3x}.$
Subtract
$y$ from each side and add
$4{x}^{2}-5\text{:}$
We now develop a solution technique for any first-order linear differential equation. We start with the standard form of a first-order linear differential equation:
$y\prime +p\left(x\right)y=q\left(x\right).$
The first term on the left-hand side of
[link] is the derivative of the unknown function, and the second term is the product of a known function with the unknown function. This is somewhat reminiscent of the power rule from the
Differentiation Rules section. If we multiply
[link] by a yet-to-be-determined function
$\mu \left(x\right),$ then the equation becomes
Matching term by term gives
$y=f\left(x\right),g\left(x\right)=\mu \left(x\right),$ and
${g}^{\prime}\left(x\right)=\mu \left(x\right)p\left(x\right).$ Taking the derivative of
$g\left(x\right)=\mu \left(x\right)$ and setting it equal to the right-hand side of
${g}^{\prime}\left(x\right)=\mu \left(x\right)p\left(x\right)$ leads to
This is a first-order, separable differential equation for
$\mu \left(x\right).$ We know
$p(x)$ because it appears in the differential equation we are solving. Separating variables and integrating yields
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Rafiq
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Rafiq
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Damian
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Leaves accumulate on the forest floor at a rate of 2 g/cm2/yr and also decompose at a rate of 90% per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor, assuming at time 0 there is no leaf litter on the ground. Does this amount approach a steady value? What is that value?
You have a cup of coffee at temperature 70°C, which you let cool 10 minutes before you pour in the same amount of milk at 1°C as in the preceding problem. How does the temperature compare to the previous cup after 10 minutes?