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Calculus 1 : Equations

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Example Questions

Example Question #4 : How To Find Solutions To Differential Equations

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

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

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

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

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

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

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

Possible Answers:

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

y = x^3 + x

$y = \sin x$

$y = \tan(x)$

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

Correct answer:

$y = \tan(x)$

Explanation:

As described in the problem, we are given

y'' = 2y + 2y^3 .

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

y'' y' = 2y y' + 2 y^3 y'

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

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

This yields:

(y')^2 = C + 2 y^2 + y^4

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

(y')^2 = 1 + 2 y^2 + y^4 = (1 + y^2)^2

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

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:

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

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

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

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

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Example Question #5 : How To Find Solutions To Differential Equations

What are all the functions y(x) such that

y''(x) = 1/x ?

Possible Answers:

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

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

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

y = k x + C for arbitrary constants k and C

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

Correct answer:

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

Explanation:

Integrating once, we get:

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

Integrating a second time gives:

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

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

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

Canceling the x's we get:

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

Defining k = 1 - C_1 gives the above form.

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Example Question #7 : Solutions To Differential Equations

The Fibonacci numbers are defined as

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

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

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

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

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

Find the Fibonacci function defined by:

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

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

whose derivatives at 0 are therefore the Fibonacci numbers.

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

f'' = f' + f

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

k^2 = k + 1 ,

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

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

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

C_+ + C_- = 0

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

C_- = - C_+

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

Hence the answer is:

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

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Example Question #8 : Solutions To Differential Equations

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

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

Possible Answers:

y^3-4y^2=9x^3-18

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

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

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

Correct answer:

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

Explanation:

The first thing we must do is rewrite the equation:

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

We can then find the integrals:

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

The integrals as as follows:

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

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

we're left with

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

We then plug in the initial condition and solve for

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

9-18=3+c

c=-12

The particular solution is then:

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

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Example Question #11 : Solutions To Differential Equations

Find f'(2) of the following equation:

f(x)=x^3+4x^2-7x-9

Possible Answers:

Correct answer:

Explanation:

First take the derivative and then solve when x=2.

f(x)=x^3+4x^2-7x-9

To find the derivative use the power rule which states when,

f(x)=x^n the derivative is $f'(x)=nx^{n-1}$ .

Therefore the derivative of our function is:

f'(x)=3x^2+8x-7

f'(2)=3(2)^2+8(2)-7=21

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Example Question #12 : Solutions To Differential Equations

Find f'(1) for the following equation:

f(x)=3x^5(2x+4)

Possible Answers:

Undefined

Correct answer:

Explanation:

To find the derivative of this function we will need to use the product rule which states to multiply the first function by the derivative of the second function and add that to the product of the second function and the derivative of the first function. In other words,

$f'(x)=h(x)\cdot g'(x)+g(x)\cdot h'(x)$

To do this we will let,

h(x)=3x^5 and g(x)=2x+4

h'(x)=15x^4 and g'(x)=2

Now we can find the derivative by plugging in these equations as follows.

f(x)=3x^5(2x+4)

f'(x)=15x^2(2x+4)+3x^5(2)

Now plug in x=1 and solve.

f'(1)=15(1)^2(2(1)+4)+3(1)^5(2)=96

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Example Question #13 : Solutions To Differential Equations

Find the solution to the following equation at f'(x)=-2

f(x)=4x^3-6x+1

Possible Answers:

Undefined

Correct answer:

Explanation:

To solve, we must first find the derivative and then solve when x=-2.

To find the derivative of the function we will use the Power Rule:

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$

Therefore,

f(x)=4x^3-6x+1

f'(x)=12x^2-6

Now to solve for -2 we plug it into our x value.

f'(-2)=12(-2)^2-6=48-6=42

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Example Question #14 : Solutions To Differential Equations

Find f'(3) for the following equation:

$f(x)=\ln(x)+7x^2-19x$

Possible Answers:

$\ln(3)+6$

$\frac{23}{3}$

$23\frac{1}{3}$

$\ln(3)+23$

Correct answer:

$23\frac{1}{3}$

Explanation:

First, find the derivative. Then, evaluate at x=3.

For this function we will use the Power Rule to find the derivative.

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$

Also remember that the derivative of $\ln(x)$ is $\frac{1}{x}$ .

Therefore we get,

$f(x)=\ln(x)+7x^2-19x$

$f'(x)=\frac{1}{x}+14x-19$

$f'(3)=\frac{1}{3}+14(3)-19=23\frac{1}{3}$

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Example Question #15 : Solutions To Differential Equations

Find the particular solution given $y ( \frac{1}{2} )= -1$ .

$\frac{\mathrm{d}y }{\mathrm{d} x}=2xy^2$

Possible Answers:

$\frac{1}{y}=x^2+\frac{5}{4}$

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

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

$-\frac{3}{y^3}=x^2+\frac{11}{4}$

Correct answer:

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

Explanation:

The first thing we must do is rewrite the equation:

$\left(\frac{1}{y^2}\right)dy=(2x) dx$

We can then find the integrals:

$\int \left(\frac{1}{y^2}\right)dy=\int (2x) dx$

The integrals are as follows:

$\int \left(\frac{1}{y^2}\right)dy= \int y^{-2} = -1y^{-1}=-\frac{1}{y}$

$\int (2x) dx=x^2$

We're left with:

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

We then plug in the initial condition and solve for

$-\frac{1}{(-1)}=\left(\frac{1}{2}\right)^2+c$

$1=\left(\frac{1}{4}\right)+c$

$c=\frac{3}{4}$

The particular solution is then:

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

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Example Question #11 : How To Find Solutions To Differential Equations

Find the particular solution given $y\left(-\frac{1}{2}\right)=-\frac{1}{4}$ .

$\frac{\mathrm{d} y}{\mathrm{d} x}=6x^2y^2$

Possible Answers:

$-\frac{1}{3y^3}=2x^3+\frac{259}{12}$

$-\frac{1}{y}=2x^3+\frac{17}{4}$

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

$-\frac{1}{y}=2x^3+\frac{1}{8}$

Correct answer:

$-\frac{1}{y}=2x^3+\frac{17}{4}$

Explanation:

The first thing we must do is rewrite the equation:

$\left(\frac{1}{y^2}\right)dy=(6x^2)dx$

We can then find the integrals:

$\int \left(\frac{1}{y^2}\right)dy=\int (6x^2)dx$

The integrals are as follows:

$\int \left(\frac{1}{y^2}\right)dy=-\frac{1}{y}$

$\int (6x^2)dx=2x^3$

We're left with:

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

We then plug in the initial condition and solve for

$-\frac{1}{(-\frac{1}{4})}=2\left(-\frac{1}{2}\right)^3+c$

$4=-\frac{1}{4}+c$

$c=\frac{17}{4}$

The particular solution is then:

$-\frac{1}{y}=2x^3+\frac{17}{4}$

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Calculus 1 : Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #4 : How To Find Solutions To Differential Equations

Example Question #5 : How To Find Solutions To Differential Equations

Example Question #7 : Solutions To Differential Equations

Example Question #8 : Solutions To Differential Equations

Example Question #11 : Solutions To Differential Equations

Example Question #12 : Solutions To Differential Equations

Example Question #13 : Solutions To Differential Equations

Example Question #14 : Solutions To Differential Equations

Example Question #15 : Solutions To Differential Equations

Example Question #11 : How To Find Solutions To Differential Equations

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