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Calculus 1 : Solutions to Differential Equations

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Example Questions

Example Question #1 : Solutions To Differential Equations

Find the derivative of (5+3x)⁵.

Possible Answers:

15(5+3x)^4

15x(5+3x)^4

5x(5+3x)^4

5(5+3x)^4x

5(5+3x)^4

Correct answer: 15(5+3x)^4

Explanation:

We'll solve this using the chain rule.

D_x[(5+3x)⁵]

=5(5+3x)⁴ * D_x[5+3x]

=5(5+3x)⁴(3)

=15(5+3x)⁴

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Example Question #1 : How To Find Solutions To Differential Equations

Find D_x[sin(7x)].

Possible Answers:

-7cos(7x)

7sin(7x)

-7sin(7x)

7cos(7x)

7sin(7x)cos(7x)

Correct answer: 7cos(7x)

Explanation:

First, remember that D_x[sin(x)]=cos(x). Now we can solve the problem using the Chain Rule.

D_x[sin(7x)]

=cos(7x)*D_x[7x]

=cos(7x)*(7)

=7cos(7x)

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Example Question #3 : Solutions To Differential Equations

Calculate f_xxyz if f(x,y,z)=sin(4x+yz).

Possible Answers:

cos(4x+yz)

-16cos(4x+yz) +16yzsin(4x+yz)

4sin(4x+yz)

-16sin(4x+yz)

arctan(4x+yz)

Correct answer: -16cos(4x+yz) +16yzsin(4x+yz)

Explanation:

We can calculate this answer in steps. We start with differentiating in terms of the left most variable in "xxyz". So here we start by taking the derivative with respect to x.

First, f_x= 4cos(4x+yz)

Then, f_xx= -16sin(4x+yz)

f_xxy= -16zcos(4x+yz)

Finally, f_xxyz= -16cos(4x+yz) + 16yzsin(4x+yz)

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Example Question #1 : How To Find Solutions To Differential Equations

Integrate $\small \int_{0}^{\pi } \cos x \hspace 2 dx$ $\displaystyle \small \int_{0}^{\pi } \cos x \hspace 2 dx$

Possible Answers:

$\displaystyle -1$

$\small \frac {\pi}{2}$ $\displaystyle \small \frac {\pi}{2}$

$\displaystyle 1$

$\displaystyle 0$

$\small -\frac{\pi}{2}$ $\displaystyle \small -\frac{\pi}{2}$

Correct answer:

$\displaystyle 0$

Explanation:

$\frac{\mathrm{d} }{\mathrm{d} x}\sin x=cosx$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\sin x=cosx$

thus:

$\small \small \int_{0}^{\pi } \cos x \hspace 2 dx = \sin x \left]^\pi_0$ $\displaystyle \small \small \int_{0}^{\pi } \cos x \hspace 2 dx = \sin x \left]^\pi_0$
$\small \sin \pi - \sin 0 = 0 - 0 = 0$ $\displaystyle \small \sin \pi - \sin 0 = 0 - 0 = 0$

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Example Question #252 : Equations

Integrate :

$\small \int_{-\frac{\pi}{4}}^{0} \sec x \tan x \hspace 2 dx$ $\displaystyle \small \int_{-\frac{\pi}{4}}^{0} \sec x \tan x \hspace 2 dx$

Possible Answers:

$\small 0$ $\displaystyle \small 0$

$\small \small \sqrt{2}$ $\displaystyle \small \small \sqrt{2}$

$\small 1+\sqrt{2}$ $\displaystyle \small 1+\sqrt{2}$

$\small \sqrt{2} -1$ $\displaystyle \small \sqrt{2} -1$

$\small 1 - \sqrt{2}$ $\displaystyle \small 1 - \sqrt{2}$

Correct answer:

$\small 1 - \sqrt{2}$ $\displaystyle \small 1 - \sqrt{2}$

Explanation:

$\frac{\mathrm{d} }{\mathrm{d} x}\sec x=\sec x\tan x$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\sec x=\sec x\tan x$

thus:
$\small \small \int_{-\frac{\pi}{4}}^{0} \sec x \tan x \hspace 2 dx = \sec x \left]^0_{-\frac{\pi}{4} }$ $\displaystyle \small \small \int_{-\frac{\pi}{4}}^{0} \sec x \tan x \hspace 2 dx = \sec x \left]^0_{-\frac{\pi}{4} }$
$\small = \sec 0 - \sec(-\frac{\pi}{4}) = 1 - \sqrt{2}$ $\displaystyle \small = \sec 0 - \sec(-\frac{\pi}{4}) = 1 - \sqrt{2}$

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Example Question #13 : Differential Equations

Find the general solution, y(x) $\displaystyle y(x)$, to the differential equation

$\frac{dy}{dx}=30-2x$ $\displaystyle \frac{dy}{dx}=30-2x$.

Possible Answers:

$\displaystyle Ln(30-x^2)$

30x-x^2 $\displaystyle 30x-x^2$

$\displaystyle -Ln(30-x)+C$

$\displaystyle 30x-x^2+C$

Correct answer:

$\displaystyle 30x-x^2+C$

Explanation:

We can use separation of variables to solve this problem since all of the "y-terms" are on one side and all of the "x-terms" are on the other side. The equation can be written as $\frac{dy}{dx}=30-2x\Rightarrow dy=(30-2x)dx$ $\displaystyle \frac{dy}{dx}=30-2x\Rightarrow dy=(30-2x)dx$.

Integrating both sides gives us $\int dy = \int (30-2x) dx\Rightarrow y(x) = 30x-x^2+C$ $\displaystyle \int dy = \int (30-2x) dx\Rightarrow y(x) = 30x-x^2+C$.

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Example Question #1 : How To Find Solutions To Differential Equations

Consider $\displaystyle y''= f(y)$ ; by multiplying by y'(x) $\displaystyle y'(x)$ both the left and the right hand sides can be swiftly integrated as

$\frac{1}{2}(y')^2 = F(y) + C$ $\displaystyle \frac{1}{2}(y')^2 = F(y) + C$

where $\displaystyle dF/dy = f(y)$. So, for example, $\displaystyle y''(x) = [y(x)]^2$ can be rewritten as:

$\frac{1}{2} [y'(x)]^2 = \frac{1}{3} [y(x)]^3 + C$ $\displaystyle \frac{1}{2} [y'(x)]^2 = \frac{1}{3} [y(x)]^3 + C$. We will use this trick on another simple case with an exact integral.

Use the technique above to find y(x) $\displaystyle y(x)$ such that $\displaystyle y''(x) = 2 y(x) + 2 [y(x)]^3$ with $\displaystyle y(0) = 0$ and $\displaystyle y'(0) = 1$.

Hint: Once you use the above to simplify the expression to the form $\displaystyle y' = g(y)$, you can solve it by moving g(y) $\displaystyle g(y)$ into the denominator:

$\int dy/g(y) = \int dx$ $\displaystyle \int dy/g(y) = \int dx$

Possible Answers:

$\displaystyle y = x^3 + x$

$y = 2 ~ \frac{e^x - 1}{e^x + 1}$ $\displaystyle y = 2 ~ \frac{e^x - 1}{e^x + 1}$

$y = \sin x$ $\displaystyle y = \sin x$

$y = \tan^{-1}(x)$ $\displaystyle y = \tan^{-1}(x)$

$y = \tan(x)$ $\displaystyle y = \tan(x)$

Correct answer:

$y = \tan(x)$ $\displaystyle y = \tan(x)$

Explanation:

As described in the problem, we are given

$\displaystyle y'' = 2y + 2y^3$.

We can multiply both sides by $\displaystyle y' = dy/dx$:

$\displaystyle y'' y' = 2y y' + 2 y^3 y'$

Recognize the pattern of the chain rule in two different ways:

$\frac{1}{2}[(y')^2]' = [y^2]' + \frac{1}{2} [y^4]'$ $\displaystyle \frac{1}{2}[(y')^2]' = [y^2]' + \frac{1}{2} [y^4]'$

This yields:

$\displaystyle (y')^2 = C + 2 y^2 + y^4$

We use the initial conditions to solve for C, noticing that at $\displaystyle x=0, y=0$ and y' = 1. $\displaystyle y' = 1.$ This means that C must be 1 above, which makes the right hand side a perfect square:

$\displaystyle (y')^2 = 1 + 2 y^2 + y^4 = (1 + y^2)^2$

$\frac{dy}{dx} = \pm (1 + y^2)$ $\displaystyle \frac{dy}{dx} = \pm (1 + y^2)$

To see whether the + or - symbol is to be used, we see that the derivative starts out positive, so the positive square root is to be used. Then following the hint we can rewrite it as:

$\int \frac{dy}{1 + y^2} = \int dx$ $\displaystyle \int \frac{dy}{1 + y^2} = \int dx$,

which we learned to solve by the trigonometric substitution$\displaystyle y = tan(u)$, yielding:

$\tan^{-1} y = x + C$ $\displaystyle \tan^{-1} y = x + C$

Clearly $\displaystyle y = tan(x + C)$ and the fact that $\displaystyle y(0) = 0$ again gives us C = 0, $\displaystyle C = 0,$ so $\displaystyle y = tan(x).$

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Example Question #2 : How To Find Solutions To Differential Equations

What are all the functions y(x) $\displaystyle y(x)$ such that

$\displaystyle y''(x) = 1/x$?

Possible Answers:

$y = e^{k \ln x} + C$ $\displaystyle y = e^{k \ln x} + C$ for arbitrary constants k and C

$y = x \ln x - k x + C$ $\displaystyle y = x \ln x - k x + C$ for arbitrary constants k and C

$\displaystyle y = k x + C$ for arbitrary constants k and C

$y = \sqrt{\ln(k x^2)} + C$ $\displaystyle y = \sqrt{\ln(k x^2)} + C$ for arbitrary constants k and C

$y = \frac{2}{x^3} + k x + C$ $\displaystyle y = \frac{2}{x^3} + k x + C$ for arbitrary constants k and C

Correct answer:

$y = x \ln x - k x + C$ $\displaystyle y = x \ln x - k x + C$ for arbitrary constants k and C

Explanation:

Integrating once, we get:

$y'(x) = \int \frac{dx}{x} + C_1 = \ln x + C_1$ $\displaystyle y'(x) = \int \frac{dx}{x} + C_1 = \ln x + C_1$

Integrating a second time gives:

$y(x) = \int \ln x ~ dx + C_1 ~ x + C$ $\displaystyle y(x) = \int \ln x ~ dx + C_1 ~ x + C$

We integrate the first term by parts using $\displaystyle u = ln(x), dv = x$ to get:

$y(x) = x ~ \ln x ~-~ \int x ~ \frac{dx}{x} ~+~ C_1 ~ x ~+~ C$ $\displaystyle y(x) = x ~ \ln x ~-~ \int x ~ \frac{dx}{x} ~+~ C_1 ~ x ~+~ C$

Canceling the x's we get:

$y(x) = x ~ \ln x ~+~ \left( C_1 - 1 \right ) x ~+~ C$ $\displaystyle y(x) = x ~ \ln x ~+~ \left( C_1 - 1 \right ) x ~+~ C$

Defining $\displaystyle k = 1 - C_1$ gives the above form.

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Example Question #1 : Solutions To Differential Equations

The Fibonacci numbers are defined as

$F_0 = 0, F_1 = 1, F_n = F_{n-1} + F_{n-2}$ $\displaystyle F_0 = 0, F_1 = 1, F_n = F_{n-1} + F_{n-2}$

and are intimately tied to the golden ratios $\phi_\pm = (1 \pm \sqrt{5})/2$ $\displaystyle \phi_\pm = (1 \pm \sqrt{5})/2$, which solve the very similar equation

$\phi^n_\pm = \phi^{n - 1}_\pm + \phi^{n-2}_\pm$ $\displaystyle \phi^n_\pm = \phi^{n - 1}_\pm + \phi^{n-2}_\pm$.

The n'th derivatives of a function are defined as:

$y^{(0)}(x) = y(x), y^{(n)}(x) = \frac{d}{dx} y^{(n - 1)}(x)$ $\displaystyle y^{(0)}(x) = y(x), y^{(n)}(x) = \frac{d}{dx} y^{(n - 1)}(x)$

Find the Fibonacci function defined by:

$f^{(n)}(x) = f^{(n-1)}(x) + f^{(n-2)}(x)$ $\displaystyle f^{(n)}(x) = f^{(n-1)}(x) + f^{(n-2)}(x)$

$f^{(0)}(0) = 0, f^{(1)}(0) = 1$ $\displaystyle f^{(0)}(0) = 0, f^{(1)}(0) = 1$

whose derivatives at 0 are therefore the Fibonacci numbers.

Possible Answers:

$f(x) = \sum_{n=0}^{\infty} \frac{\phi_+^n}{n!} ~ x^n$ $\displaystyle f(x) = \sum_{n=0}^{\infty} \frac{\phi_+^n}{n!} ~ x^n$

$f(x) = \sum_{n=0}^{\infty} F_n ~ x^n$ $\displaystyle f(x) = \sum_{n=0}^{\infty} F_n ~ x^n$

$f(x) = e^{x/2} \sin(x \sqrt{3/4}) / \sqrt{3/4}$ $\displaystyle f(x) = e^{x/2} \sin(x \sqrt{3/4}) / \sqrt{3/4}$

$f(x) = (e^{x \phi_+} - e^{x \phi_-})/\sqrt{5}$ $\displaystyle f(x) = (e^{x \phi_+} - e^{x \phi_-})/\sqrt{5}$

$f(x) = (\phi_+^x - \phi_-^x) / 2$ $\displaystyle f(x) = (\phi_+^x - \phi_-^x) / 2$

Correct answer:

$f(x) = (e^{x \phi_+} - e^{x \phi_-})/\sqrt{5}$ $\displaystyle f(x) = (e^{x \phi_+} - e^{x \phi_-})/\sqrt{5}$

Explanation:

To solve $f^{(n)} = f^{(n-1)} + f^{(n-2)}$ $\displaystyle f^{(n)} = f^{(n-1)} + f^{(n-2)}$, we ignore n-2 $\displaystyle n-2$ of the derivatives to get simply:

$\displaystyle f'' = f' + f$

This can be solved by assuming an exponential function $y = C e^{k x}$ $\displaystyle y = C e^{k x}$, which turns this expression into

$\displaystyle k^2 = k + 1$,

which is solved by $k = \phi_\pm$ $\displaystyle k = \phi_\pm$ . Our general solution must take the form:

$f(x) = C_+ ~ e^{x \phi_+} + C_- ~ e^{x \phi_-}$ $\displaystyle f(x) = C_+ ~ e^{x \phi_+} + C_- ~ e^{x \phi_-}$

Plugging in our initial conditions $\displaystyle f(0) = 0$ and $\displaystyle f'(0) = 1$, we get:

$\displaystyle C_+ + C_- = 0$

$C_+ \ \phi_+ ~+~ C_- \ \phi_- = 1$ $\displaystyle C_+ \ \phi_+ ~+~ C_- \ \phi_- = 1$

$\displaystyle C_- = - C_+$

$C_+ = 1/(\phi_+ - \phi_-) = 1/(\frac{1+\sqrt 5}{2} - \frac{1-\sqrt 5}{2}) = 1/\sqrt{5}$ $\displaystyle C_+ = 1/(\phi_+ - \phi_-) = 1/(\frac{1+\sqrt 5}{2} - \frac{1-\sqrt 5}{2}) = 1/\sqrt{5}$

Hence the answer is:

$f(x) = (e^{x \phi_+} - e^{x \phi_-}) / \sqrt{5}$ $\displaystyle f(x) = (e^{x \phi_+} - e^{x \phi_-}) / \sqrt{5}$

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Example Question #2 : Solutions To Differential Equations

Find the particular solution given y(1)=3 $\displaystyle y(1)=3$.

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{9x^2}{y^2-4y}$ $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{9x^2}{y^2-4y}$

Possible Answers:

$\frac{1}{3}y^3-2y^2=3x^3+12$ $\displaystyle \frac{1}{3}y^3-2y^2=3x^3+12$

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

$\displaystyle y^3-4y^2=9x^3-18$

$\frac{1}{3}y^3-2y^2=3x^3-12$ $\displaystyle \frac{1}{3}y^3-2y^2=3x^3-12$

Correct answer:

$\frac{1}{3}y^3-2y^2=3x^3-12$ $\displaystyle \frac{1}{3}y^3-2y^2=3x^3-12$

Explanation:

The first thing we must do is rewrite the equation:

$\displaystyle (y^2-4y)dy=(9x^2)dx$

We can then find the integrals:

$\int (y^2-4y)dy= \int (9x^2)dx$ $\displaystyle \int (y^2-4y)dy= \int (9x^2)dx$

The integrals as as follows:

$\int (y^2-4y)dy= \frac{1}{3}y^3-2y^2$ $\displaystyle \int (y^2-4y)dy= \frac{1}{3}y^3-2y^2$

$\int (9x^2)dx=3x^3$ $\displaystyle \int (9x^2)dx=3x^3$

we're left with

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

We then plug in the initial condition and solve for $\displaystyle c.$

$\frac{1}{3}(3)^3-2(3)^2=3(1)^3+c$ $\displaystyle \frac{1}{3}(3)^3-2(3)^2=3(1)^3+c$

$\displaystyle 9-18=3+c$

c=-12 $\displaystyle c=-12$

The particular solution is then:

$\frac{1}{3}y^3-2y^2=3x^3-12$ $\displaystyle \frac{1}{3}y^3-2y^2=3x^3-12$

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Calculus 1 : Solutions to Differential Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1 : Solutions To Differential Equations

Example Question #1 : How To Find Solutions To Differential Equations

Example Question #3 : Solutions To Differential Equations

Example Question #1 : How To Find Solutions To Differential Equations

Example Question #252 : Equations

Example Question #13 : Differential Equations

Example Question #1 : How To Find Solutions To Differential Equations

Example Question #2 : How To Find Solutions To Differential Equations

Example Question #1 : Solutions To Differential Equations

Example Question #2 : Solutions To Differential Equations

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