We know \( y = 5e^{x^4} \) so \( \frac{dy}{dx} = 20x^3e^{x^4} = 20x^3y \). (Chain rule used)
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 \( c=5 \).
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.
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*}
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 |
\[ Error = \frac{Actual\:value - Expected\:value}{Expected\:value}\cdot 100 = \frac{0.641-0.515}{0.515} \cdot 100 = 24.5\% \]
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:
\[y = \frac{1}{\sqrt{-3x^2 + 4x - \frac{3}{4}}}\]
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:
\[z = \frac{1}{\sqrt{\frac{63x + 1}{4}}} = \frac{2}{\sqrt{63x + 1}}\]
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 \( y = vx \):
\[y = x \left( \frac{\sqrt{3}}{2} \tan\left(\frac{\sqrt{3}}{2} \ln|x| - \frac{\pi}{6}\right) - \frac{3}{2} \right)\]
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:
\[y = x \sin x\]
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 \( x = 0.3 \):
| \( n \) | \( x_n \) | \( y_n \) | \( f(x_n, y_n) = e^{-x_n} - 3y_n^2 \) |
|---|---|---|---|
| \( 0 \) | \( 0.0 \) | \( 2.000 \) | \( e^{0} - 3(2)^2 = 1 - 12 = -11 \) |
| \( 1 \) | \( 0.1 \) | \( 2.000 + 0.1(-11) = 0.900 \) | \( e^{-0.1} - 3(0.9)^2 \approx 0.9048 - 2.43 = -1.5252 \) |
| \( 2 \) | \( 0.2 \) | \( 0.900 + 0.1(-1.5252) = 0.7475 \) | \( e^{-0.2} - 3(0.7475)^2 \approx 0.8187 - 1.6757 = -0.8570 \) |
| \( 3 \) | \( 0.3 \) | \( 0.7475 + 0.1(-0.8570) = 0.6618 \) |
The approximate value of \( y \) when \( x = 0.3 \) is:
\[ y_3 = 0.6618 \]
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 \( x = 0.4 \):
| \( n \) | \( x_n \) | \( y_n \) | \( f(x_n, y_n) = e^{x_n} - 4y_n^2 \) |
|---|---|---|---|
| \( 0 \) | \( 0.0 \) | \( 1.000 \) | \( e^{0} - 4(1)^2 = 1 - 4 = -3 \) |
| \( 1 \) | \( 0.1 \) | \( 1.000 + 0.1(-3) = 0.700 \) | \( e^{0.1} - 4(0.7)^2 \approx 1.1052 - 1.96 = -0.8548 \) |
| \( 2 \) | \( 0.2 \) | \( 0.700 + 0.1(-0.8548) = 0.6145 \) | \( e^{0.2} - 4(0.6145)^2 \approx 1.2214 - 1.5098 = -0.2884 \) |
| \( 3 \) | \( 0.3 \) | \( 0.6145 + 0.1(-0.2884) = 0.5857 \) | \( e^{0.3} - 4(0.5857)^2 \approx 1.3499 - 1.3722 = -0.0223 \) |
| \( 4 \) | \( 0.4 \) | \( 0.5857 + 0.1(-0.0223) = 0.5835 \) |
The approximate value of \( y \) when \( x = 0.4 \) is:
\[ y_4 = 0.5835 \]
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 \( y \):
\[ y = \frac{x+3}{x+2} \left( \ln|x+3| + \frac{1}{x+3} + C \right) \]