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Calculus 1 : Differential Equations

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Example Questions

Example Question #2444 : Calculus

Find the explicit function of given

$\frac{dy}{dx}=xy$ .

Possible Answers:

$y=Ce^{\frac{x^2}{2}}$

y=e^2+c

y=Ce^x

$y=e^{x^2}$

Correct answer:

$y=Ce^{\frac{x^2}{2}}$

Explanation:

In order to determine the explicit function of y, we must separate the variables onto each side of the equation

becomes

$\frac{dy}{y}=x\,dx$

Integrating both sides of the equation

and by applying the inverse power rule for the right-hand-side which says

$\int x^n = \frac{x^{n+1}}{n+1}$

yields

$ln|y|=\frac{x^2}{2}+c$

Exponentiating both sides of the equation, we obtain

$y=Ce^{\frac{x^2}{2}}$

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Example Question #112 : How To Find Solutions To Differential Equations

Find the derivative of $f(s) = 2se^{2s}$ .

Possible Answers:

$2e^{2s} + 2se^{2s}$

$2e^{2s}$

$2se^{2s}$

$2e^{2s} + 4se^{2s}$

$4e^{2s}$

Correct answer:

$2e^{2s} + 4se^{2s}$

Explanation:

This function is composed of two functions multiplied together; therefore you must use the product rule to find the derivative. The product rule is given by:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x) \cdot g(x) = f'g + g'f$

Note in this case the two functions:

f(x) = 2s

$g(x) = e^{2s}$

The derivatives are:

f'(x) = 2

$g'(x) = 2e^{2s}$

Using the derivatives of the two functions and applying the product rule, you will recieve the proper derivative:

$2e^{2s} + 4se^{2s}$

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Example Question #113 : How To Find Solutions To Differential Equations

Which differential equations does y=ln(x) solve? Assume x>0 .

Possible Answers:

y'=e^y

$y'=\frac{1}{y}$

$y'=\frac{1}{e^y}$

$y=\frac{1}{e^y'}$

Correct answer:

$y'=\frac{1}{e^y}$

Explanation:

You can solve this equation by plugging in y=ln(x) into each answer choice and seeing if the two sides are equal.

Another method is to arrange in terms of itself and its derivative.

y=ln(x) , x>0

$y'=\frac{1}{x}$ . This is an identity.

We want to create a differential equation that equates those two terms.

y=ln(x)

Raising both sides by

$e^y=e^{ln(x)}$ .

Recall that $e^{ln(x)}=x$

e^y=x .

Raising both sides to the negative first power,

$(e^y)^{-1}=x^{-1}$

$\frac{1}{e^y}=\frac{1}{x}$

Recall that $\frac{1}{x}=y'$

Therefore,

$y'=\frac{1}{e^y}$

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Example Question #364 : Equations

Find the general solution to the differential equation given by:

y''=3t

Assume is a function of . and given below are constants.

Possible Answers:

$y=\frac{1}{2}t^3+Ct$

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

$y=\frac{1}{2}t^3+Bt+C$

y=0

Correct answer:

$y=\frac{1}{2}t^3+Bt+C$

Explanation:

To solve this, we only have to take the integral of both sides twice, and that will remove the y'' term.

Remember the power rule for when we do integration on polynomials.

By the power rule, we know that

$\int at^n dt= \frac{a}{n+1}t^{(n+1)} +C$ , where a, n, C are constants and is a variable.

$\int y''dt=\int (3t)dt$

$y'=\frac{3}{2}t^2+B$ , where is a constant

$\int y'dt=\int (\frac{3}{2}t^2+B)dt$

$y=\frac{1}{2}t^3+Bt+C$ , where and are constants. This is the most correct notation.

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Example Question #115 : How To Find Solutions To Differential Equations

In what interval(s) is the graph of the function $f(x)=x^2 + \frac{1}{x}$ concave down?

Possible Answers:

$(-\frac{\sqrt2}{2}, \frac{\sqrt2}{2})$

$(-\infty, 0)$

$(-\infty, 0) \cup (0, \infty)$

$(-\infty, -\frac{\sqrt2}{2}) \cup (\frac{\sqrt2}{2}, \infty)$

Never (The graph is always concave up.)

Correct answer:

$(-\frac{\sqrt2}{2}, \frac{\sqrt2}{2})$

Explanation:

The function is concave down when f''(x) < 0. $f'(x) = 2x+ \frac{1}{x} \Rightarrow f''(x) = 2-\frac{1}{x^2}.$

$f''(x)<0 \Rightarrow 2<\frac{1}{x^2} \Rightarrow 2x^2<1 \Rightarrow x^2< \frac{1}{2}. \\ \Rightarrow x \in (- \frac{\sqrt2}{2}, \frac{\sqrt2}{2}).$

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Example Question #2441 : Calculus

Find the general solution of the following differential equation:

$\frac{\mathrm{dy} }{\mathrm{d} x}=3y\sec^2(x)$

Possible Answers:

$y=Ce^{3\tan(x)}$

$y=Ce^{3\sec(x)tan(x)}$

y=x

$y=e^{3\tan(x)}+C$

Correct answer:

$y=Ce^{3\tan(x)}$

Explanation:

To find the general solution for the separable differential equation, we must move x and dx, y and dy to separate sides, and then integrate both sides:

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

Next, integrate both sides:

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

The rules used for the integrations are:

$\int \frac{dx}{x}=\ln\left | x\right |+C$ , $\int \sec^2(x)dx=\tan(x)+C$

Note that both Cs were combined to make one constant of integration in our equation.

Finally, solve for y:

$y=Ce^{3\tan(x)}$

Note that C was brought to the front, as e^C is itself a constant of integration.

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Example Question #2451 : Calculus

Derivative Rules

What is the derivative power rule?

Possible Answers:

For positive integers n, $\frac{d}{dx}x^{n}=(n-1)x^{n}$

For positive integers n, $\frac{d}{dx}x^{n}=nx^{n-1}$

For positive integers e, $\frac{d}{dx}e^{x}=e^x$

For positive integers n, $\frac{d}{dx}x^{n}=nx^{1-n}$

For all integers n, $\frac{d}{dx}x^{n}=nx^{n-1}$

Correct answer:

For positive integers n, $\frac{d}{dx}x^{n}=nx^{n-1}$

Explanation:

\frac{d}{dx} x^n = nx^{n-1} , \qquad n \neq 0.

The power rule holds for all powers except for the constant value $x^0$

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Example Question #121 : Differential Equations

Find f'(1) given:

f(x)=-3x^2-4x+8

Possible Answers:

Correct answer:

Explanation:

To solve, take the first derivative and evaluate at . Thus,

f'(x)=-6x-4

f'(1)=-6(1)-4=-10

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Example Question #1422 : Functions

Solve the initial value problem: y' +3y = 1, y(0) =1 .

Possible Answers:

$y = \frac{1}{3}+\frac{1}{2e^{3t}}$

$y = \frac{1}{3}+\frac{2}{3e^{3t}}$

$y = \frac{1}{3}+\frac{1}{3e^{3t}}$

$y = \frac{2}{3}+\frac{2}{3e^{3t}}$

$y = \frac{1}{3}+\frac{3}{2e^{3t}}$

Correct answer:

$y = \frac{1}{3}+\frac{2}{3e^{3t}}$

Explanation:

The differential equation is in its correct form.

$\frac{dy}{dx} +p(t)y = g(t)$

Solve for the integrating factor.

$u(t) = e^{\int p(t)dt}$

$u(t) = e^{\int 3dt} = e^{3t}$

Multiply the integration factor throughout the entire equation.

$e^{3t}[y' +3y = 1]$

$e^{3t}y' +3e^{3t}y = e^{3t}$

The left side of the equation becomes $(e^{3t}y)'$ from the use of our integrating factor. Rewrite the equation.

$(e^{3t}y)'=e^{3t}$

Integrate both sides.

$\int (e^{3t}y)' \:dt=\int e^{3t}\:dt$

$e^{3t}y+C_1= \frac{1}{3}e^{3t}+C_2$

Merge the constants.

$e^{3t}y = \frac{1}{3}e^{3t}+C$

Divide by $e^{3t}$ on both sides.

$y = \frac{1}{3}+\frac{C}{e^{3t}}$

Substitute the initial condition y(0) =1 to solve for .

$1 = \frac{1}{3}+\frac{C}{e^{3(0)}}$

$\frac{2}{3}=\frac{C}{e^{3(0)}}$

$\frac{2e^{3(0)}}{3}=C$

$C=\frac{2}{3}$

Resubstitute the constant. The answer is:

$y = \frac{1}{3}+\frac{2}{3e^{3t}}$

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Example Question #122 : Differential Equations

Find f'(-1) given:

f(x)=x^2-16x+11

Possible Answers:

Correct answer:

Explanation:

To solve, simply find the first derivative and let x=-1 . Thus,

$f'(x)=2x-16\Rightarrow f'(-1)=2(-1)-16=-2-16=-18$

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Calculus 1 : Differential Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2444 : Calculus

Example Question #112 : How To Find Solutions To Differential Equations

Example Question #113 : How To Find Solutions To Differential Equations

Example Question #364 : Equations

Example Question #115 : How To Find Solutions To Differential Equations

Example Question #2441 : Calculus

Example Question #2451 : Calculus

Example Question #121 : Differential Equations

Example Question #1422 : Functions

Example Question #122 : Differential Equations

All Calculus 1 Resources

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