Detailed Answer

We know $y = 5e^{x^4}$ so $\frac{dy}{dx} = 20x^3e^{x^4} = 20x^3y$. (Chain rule used)

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Detailed Answer

a) If $P=ce^{\frac{t}{2}}$, then $\frac{dP}{dt} = \frac{c}{2}e^{\frac{t}{2}} = \frac{P}{2}$.

b) The population of alligators cannot be negative.

c) We know when $t=0$, $P=5$, so plugging into the equation in (a), we find $c=5$.

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Detailed Answer

a) $y = \int 6x^2 \, dx = 2x^3 + C$

b) $y = \int \frac{1}{2x+6} \, dx = \frac{1}{2}\ln{(2x+6)} + C$

c) $y = \int \frac{p}{\sqrt{36-p^2}} \, dp$
We can do a u-substitution here, letting $u = 36-p^2$, so $\frac{du}{dp}=-2p$, and substituting these in:

$$\begin{align*} y &= \int -\frac{1}{2}\frac{1}{\sqrt{u}} \, du\\ y &= -\frac{1}{2}\left(2\sqrt{u} + C\right)\\ y &= -\sqrt{36-p^2}+C \end{align*}$$

d) This is a separable differential equation:

$$\begin{align*} \frac{dy}{y-1} &= \frac{dx}{x+5} \\ \int \frac{dy}{y-1} &= \int \frac{dx}{x+5} \\ \ln{|y-1|} &= \ln{|x+5|} + C\\ \ln{|y-1|} - \ln{|x+5|} &= C \\ \ln{\left|\frac{y-1}{x+5}\right|} &= C \\ \frac{y-1}{x+5} &= e^C \\ y &= e^C(x+5) + 1 \end{align*}$$

Note: It is common practice to replace $e^C$ with another constant, $A$, for aesthetics.

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Detailed Answer

We are told $\frac{dA}{dt} = k\sqrt[3]{A}$. From here, we can solve the differential equation to get the desired function:

$$\begin{align*} \frac{1}{\sqrt[3]{A}} dA &= k \, dt \\ \int \frac{dA}{\sqrt[3]{A}} &= \int k \, dt \\ \frac{3}{2} \sqrt[3]{A^2} &= kt + C \end{align*}$$

We were told that when $t=0$, the area was $2\sqrt{2}$:

$$\begin{align*} \frac{3}{2} \sqrt[3]{(2\sqrt{2})^2} &= k(0) + C \\ C &= 3 \end{align*}$$

When $t=4$, we know the area doubled, so it became $4\sqrt{2}$:

$$\begin{align*} \frac{3}{2} \sqrt[3]{(4\sqrt{2})^2} &= k(4) + 3 \\ k &= \frac{-3 + 3\sqrt[3]{4}}{4} \end{align*}$$

So finally:

$$\begin{align*} \frac{3}{2} \sqrt[3]{A^2} &= \left(\frac{-3 + 3\sqrt[3]{4}}{4}\right)t + 3 \\ \frac{1}{2} \sqrt[3]{A^2} &= \left(\frac{-1 + 1\sqrt[3]{4}}{4}\right)t + 1 \\ A(t) &= \sqrt{\left( \left(\frac{-1 + 1\sqrt[3]{4}}{2}\right)t + 2 \right)^3} \end{align*}$$

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Detailed Answer

The easiest way to solve this is by creating a table:

$$x_{n+1} = x_n + 0.25$$

$$y_{n+1} = y_n + 0.25(\frac{x_n^2+y_n^2}{2x_n^2})$$

n	$x_n$	$y_n$
0	1	0
1	1.25	0.125
2	1.50	0.251
3	1.75	0.379
4	2	0.641

$$Error = \frac{Actual\:value - Expected\:value}{Expected\:value}\cdot 100 = \frac{0.641-0.515}{0.515} \cdot 100 = 24.5\%$$

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Detailed Answer

a) We are given the differential equation:

$$\frac{dy}{dx} = (3x - 2)y^3$$

This is a separable differential equation. To solve it, we separate the variables and integrate:

$$\frac{dy}{y^3} = (3x - 2) \, dx$$

Integrate both sides:

$$\int y^{-3} \, dy = \int (3x - 2) \, dx$$

The left-hand side integrates to:

$$\int y^{-3} \, dy = -\frac{1}{2} y^{-2} + C_1$$

And the right-hand side integrates to:

$$\int (3x - 2) \, dx = \frac{3}{2}x^2 - 2x + C_2$$

Combining the results and solving for $y$:

$$-\frac{1}{2} y^{-2} = \frac{3}{2}x^2 - 2x + C$$

where $C = C_2 - C_1$ is the constant of integration. Rearranging:

$$y^{-2} = -3x^2 + 4x - 2C$$

Taking the reciprocal and square root:

$$y = \frac{1}{\sqrt{-3x^2 + 4x + D}}$$

where $D = -2C$ is a new constant.

b) Given $y(1) = 2$, substitute $x = 1$ and $y = 2$ into the general solution:

$$2 = \frac{1}{\sqrt{-3(1)^2 + 4(1) + D}}$$

Simplify:

$$2 = \frac{1}{\sqrt{-3 + 4 + D}} \implies 2 = \frac{1}{\sqrt{1 + D}}$$

Square both sides:

$$4 = \frac{1}{1 + D} \implies 1 + D = \frac{1}{4} \implies D = -\frac{3}{4}$$

Substitute $D = -\frac{3}{4}$ back into the general solution:

$$y = \frac{1}{\sqrt{-3x^2 + 4x - \frac{3}{4}}}$$

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Detailed Answer

a) The rate of change of $z$ is proportional to the cube of $z$. This can be expressed as:

$$\frac{dz}{dx} = k z^3$$

where $k$ is the constant of proportionality.

b) The differential equation is separable. To solve it, we separate the variables and integrate:

$$\frac{dz}{z^3} = k \, dx$$

Integrate both sides:

$$\int z^{-3} \, dz = \int k \, dx$$

The left-hand side integrates to:

$$\int z^{-3} \, dz = -\frac{1}{2} z^{-2} + C_1$$

And the right-hand side integrates to:

$$\int k \, dx = kx + C_2$$

Combining the results:

$$-\frac{1}{2} z^{-2} = kx + C$$

where $C = C_2 - C_1$ is the constant of integration. Rearranging:

$$z^{-2} = -2kx - 2C$$

Taking the reciprocal and square root:

$$z = \frac{1}{\sqrt{-2kx + D}}$$

where $D = -2C$ is a new constant.

c) Given $z(0) = 2$ and $z(1) = \frac{1}{4}$, we substitute these conditions into the general solution to find $k$ and $D$.

1. Substitute $x = 0$ and $z = 2$:

$$2 = \frac{1}{\sqrt{D}} \implies \sqrt{D} = \frac{1}{2} \implies D = \frac{1}{4}$$

2. Substitute $x = 1$, $z = \frac{1}{4}$, and $D = \frac{1}{4}$:

$$\frac{1}{4} = \frac{1}{\sqrt{-2k(1) + \frac{1}{4}}}$$

Simplify:

$$\frac{1}{4} = \frac{1}{\sqrt{-2k + \frac{1}{4}}} \implies \sqrt{-2k + \frac{1}{4}} = 4$$

Square both sides:

$$-2k + \frac{1}{4} = 16 \implies -2k = \frac{63}{4} \implies k = -\frac{63}{8}$$

Substitute $k = -\frac{63}{8}$ and $D = \frac{1}{4}$ into the general solution:

$$z = \frac{1}{\sqrt{-2\left(-\frac{63}{8}\right)x + \frac{1}{4}}} = \frac{1}{\sqrt{\frac{63}{4}x + \frac{1}{4}}}$$

Simplify:

$$z = \frac{1}{\sqrt{\frac{63x + 1}{4}}} = \frac{2}{\sqrt{63x + 1}}$$

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Detailed Answer

The given differential equation is:

$$x^2y' = y^2 + 4xy + 3x^2$$

Divide both sides by $x^2$ to simplify:

$$y' = \frac{y^2}{x^2} + \frac{4y}{x} + 3$$

Substitute $v = \frac{y}{x}$, then $y = vx$ and $y' = v + xv'$. Substitute these into the equation:

$$v + xv' = v^2 + 4v + 3$$

Rearrange the equation to separate $v$ and $x$:

$$xv' = v^2 + 3v + 3$$

Separate the variables:

$$\frac{dv}{v^2 + 3v + 3} = \frac{dx}{x}$$

Integrate both sides:

$$\int \frac{dv}{v^2 + 3v + 3} = \int \frac{dx}{x}$$

Complete the square in the denominator:

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

Thus, the integral becomes:

$$\int \frac{dv}{\left(v + \frac{3}{2}\right)^2 + \frac{3}{4}} = \ln|x| + C$$

Using the substitution $u = v + \frac{3}{2}$, the integral evaluates to:

$$\frac{2}{\sqrt{3}} \arctan\left(\frac{2v + 3}{\sqrt{3}}\right) = \ln|x| + C$$

Apply the initial condition: Given $y(1) = -2$, substitute $x = 1$ and $y = -2$ into $v = \frac{y}{x}$:

$$v = \frac{-2}{1} = -2$$

Substitute $v = -2$ and $x = 1$ into the integrated equation:

$$\frac{2}{\sqrt{3}} \arctan\left(\frac{2(-2) + 3}{\sqrt{3}}\right) = \ln|1| + C$$

Simplify:

$$\frac{2}{\sqrt{3}} \arctan\left(\frac{-1}{\sqrt{3}}\right) = C$$

$$C = \frac{2}{\sqrt{3}} \cdot \left(-\frac{\pi}{6}\right) = -\frac{\pi}{3\sqrt{3}}$$

Solve for $v$ and then $y$:

The equation becomes:

$$\frac{2}{\sqrt{3}} \arctan\left(\frac{2v + 3}{\sqrt{3}}\right) = \ln|x| - \frac{\pi}{3\sqrt{3}}$$

Solve for $v$:

$$\arctan\left(\frac{2v + 3}{\sqrt{3}}\right) = \frac{\sqrt{3}}{2} \ln|x| - \frac{\pi}{6}$$

Take the tangent of both sides:

$$\frac{2v + 3}{\sqrt{3}} = \tan\left(\frac{\sqrt{3}}{2} \ln|x| - \frac{\pi}{6}\right)$$

Solve for $v$:

$$v = \frac{\sqrt{3}}{2} \tan\left(\frac{\sqrt{3}}{2} \ln|x| - \frac{\pi}{6}\right) - \frac{3}{2}$$

Recall that $y = vx$:

$$y = x \left( \frac{\sqrt{3}}{2} \tan\left(\frac{\sqrt{3}}{2} \ln|x| - \frac{\pi}{6}\right) - \frac{3}{2} \right)$$

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Detailed Answer

Rewrite the differential equation:

Divide both sides by $x$:

$$y' = \frac{y}{x} + x \cos x$$

Identify the integrating factor:

The equation is linear in $y$. The integrating factor $\mu(x)$ is:

$$\mu(x) = e^{\int -\frac{1}{x} \, dx} = e^{-\ln|x|} = \frac{1}{x}$$

Multiply through by the integrating factor:

$$\frac{1}{x} y' - \frac{1}{x^2} y = \cos x$$

Recognize the left-hand side as a derivative:

$$\frac{d}{dx}\left(\frac{y}{x}\right) = \cos x$$

Integrate both sides:

$$\frac{y}{x} = \int \cos x \, dx = \sin x + C$$

Solve for $y$:

$$y = x \sin x + Cx$$

Apply the initial condition $y(\pi) = 0$:

$$0 = \pi \sin \pi + C\pi$$

$$0 = 0 + C\pi$$

$$C = 0$$

Final solution:

$$y = x \sin x$$

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Detailed Answer

The given differential equation is:

$$\frac{dy}{dx} = e^{-x} - 3y^2$$

Euler's method is given by:

$$y_{n+1} = y_n + h \cdot f(x_n, y_n)$$

where $f(x, y) = e^{-x} - 3y^2$ and $h = 0.1$.

Given $y(0) = 2$, we start with $x_0 = 0$ and $y_0 = 2$.

We compute the values step-by-step until $x = 0.3$:

The approximate value of $y$ when $x = 0.3$ is:

$$y_3 = 0.6618$$

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Detailed Answer

The given differential equation is:

$$\frac{dy}{dx} = e^{x} - 4y^2$$

Euler's method is given by:

$$y_{n+1} = y_n + h \cdot f(x_n, y_n)$$

where $f(x, y) = e^{x} - 4y^2$ and $h = 0.1$.

Given $y(0) = 1$, we start with $x_0 = 0$ and $y_0 = 1$.

We compute the values step-by-step until $x = 0.4$:

The approximate value of $y$ when $x = 0.4$ is:

$$y_4 = 0.5835$$

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Detailed Answer

a) We want to express $\frac{1}{(x+2)(x+3)}$ in partial fractions. Assume:

$$\frac{1}{(x+2)(x+3)} = \frac{A}{x+2} + \frac{B}{x+3}$$

Multiply both sides by $(x+2)(x+3)$:

$$1 = A(x+3) + B(x+2)$$

Expanding and collecting like terms:

$$1 = (A + B)x + (3A + 2B)$$

Equating the coefficients of corresponding powers of $x$:

$$\begin{cases} A + B = 0, \\ 3A + 2B = 1 \end{cases}$$

From the first equation, $A = -B$. Substituting into the second equation:

$$3(-B) + 2B = 1 \implies -3B + 2B = 1 \implies -B = 1 \implies B = -1$$

Thus, $A = 1$. The partial fraction decomposition is:

$$\frac{1}{(x+2)(x+3)} = \frac{1}{x+2} - \frac{1}{x+3}$$

b) The given differential equation is:

$$(x+2)(x+3)\frac{dy}{dx} + y = x + 2$$

Divide both sides by $(x+2)(x+3)$:

$$\frac{dy}{dx} + \frac{y}{(x+2)(x+3)} = \frac{x + 2}{(x+2)(x+3)}$$

Simplify the right-hand side:

$$\frac{x + 2}{(x+2)(x+3)} = \frac{1}{x+3}$$

Thus, the equation becomes:

$$\frac{dy}{dx} + \frac{y}{(x+2)(x+3)} = \frac{1}{x+3}$$

The integrating factor $\mu(x)$ is:

$$\mu(x) = e^{\int \frac{1}{(x+2)(x+3)} \, dx}$$

Using the partial fractions from part (a):

$$\int \frac{1}{(x+2)(x+3)} \, dx = \int \left( \frac{1}{x+2} - \frac{1}{x+3} \right) dx = \ln|x+2| - \ln|x+3|$$

Thus, the integrating factor simplifies to:

$$\mu(x) = e^{\ln|x+2| - \ln|x+3|} = \frac{x+2}{x+3}$$

Multiply both sides by $\mu(x)$:

$$\frac{x+2}{x+3} \frac{dy}{dx} + \frac{1}{x+3} y = \frac{x+2}{(x+3)^2}$$

The left-hand side is the derivative of $y \cdot \frac{x+2}{x+3}$:

$$\frac{d}{dx}\left( y \cdot \frac{x+2}{x+3} \right) = \frac{x+2}{x+3}.$$

Integrating both sides:

$$y \cdot \frac{x+2}{x+3} = \int \frac{x+2}{(x+3)^2} \, dx$$

Using substitution $u = x+3$, we get:

$$\int \frac{u - 1}{u^2} \, du = \int \left( \frac{1}{u} - \frac{1}{u^2} \right) du = \ln|u| + \frac{1}{u} + C$$

Substituting back $u = x + 3$:

$$\int \frac{x+2}{x+3} \, dx = \ln|x+3| + \frac{1}{x+3} + C$$

Thus, the equation simplifies to:

$$y \cdot \frac{x+2}{x+3} = \ln|x+3| + \frac{1}{x+3} + C$$

Solving for $y$:

$$y = \frac{x+3}{x+2} \left( \ln|x+3| + \frac{1}{x+3} + C \right)$$

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