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Calculus 1 : Equations

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Example Questions

Example Question #32 : Differential Equations

Differentiate the polynomial.

$x^{4}+3x^{3}+4x$

Possible Answers:

$3x^{3}+4x^{2}$

$4x^{3}+9x^{2}+4$

$4x^{3}+9x^{2}$

$x^{3}+3x^{2}+4$

x^2+4x+7

Correct answer:

$4x^{3}+9x^{2}+4$

Explanation:

Using the power rule, we can differentiate our first term reducing the power by one and multiplying our term by the original power. $x^{4}$ , will thus become $4x^{3}$ . The second term $3x^{3}$ , will thus become $9x^{2}$ . The last term is , will reduce to .

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Example Question #37 : Differential Equations

Find {f}'(x) .

$f(x) = \frac{e^x}{x^2}$

Possible Answers:

$\frac{x^2-e^x}{x^2}$

$\frac{e^x (x-2)}{x^3}$

$\frac{e^{x^{2}} (x-2)}{x^3}$

$\frac{(x-2)}{x^2}$

$\frac{(x-2)}{x^3}$

Correct answer:

$\frac{e^x (x-2)}{x^3}$

Explanation:

According to the quotient rule, the derivative of ,

$\frac{f(x)}{g(x)} = \frac{{f}'(x)g(x) - f(x){g}'(x))}{(g(x))^{2}}$ .

We will let $f(x) = e^{x} , f{}'(x) = e^{x} , g(x) = x^{2}$ and g{}'(x) = 2x
Plugging all of our values into the quotient rule formula we come to a final solution of :

$\frac{e^x (x-2)}{x^3}$ .

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Example Question #281 : Equations

Differentiate the polynomial.

$x^{3}+3x^{2}$

Possible Answers:

$3x^{2}+6x$

$3x^{4}+6x^{3}$

$x^{3}+3$

$3x^{3}+6x$

Correct answer:

$3x^{2}+6x$

Explanation:

Using the power rule, we can differentiate our first term reducing the power by one and multiplying our term by the original power. $x^{3}$ , will thus become $3x^{2}$ . The second term $3x^{2}$ , will thus become .

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Example Question #33 : Differential Equations

Solve the differential equation: $\frac{dy}{dx}= \frac{y}{2}$

Possible Answers:

y=Ce^0^.^5^x

y=Ce^ -^0^.^5^x

y=Ce^2^x

y=Ce^ -^0^.^2^5^x

y=Ce^x

Correct answer:

y=Ce^0^.^5^x

Explanation:

Rewrite $\frac{dy}{dx}= \frac{y}{2}$ by multiply the on both sides, and dividing on both sides of the equation.

$\frac{1}{y}dy= \frac{1}{2}dx$

Integrate both sides of the equation and solve for y.

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

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

$\left | y\right |= e^0^.^5^xe^C$

y=Ce^0^.^5^x

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Example Question #41 : Differential Equations

Find the equilibrium values for the following differential equation: y^2 + y'=y

Possible Answers:

0,1

-1,1

Correct answer:

0,1

Explanation:

To find the equilibrium values, substitute y'=0 and solve for . The equilibrium values are the solutions of the differential equation in constant form.

y^2 + y'=y

y^2 + 0=y

y^2-y=0

y(y-1)=0

y=0,1

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Example Question #42 : Differential Equations

Suppose is greater than zero. Solve the differential equation: $y'= \frac{x^2}{y^\frac{1}{2}}$

Possible Answers:

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

$y=(2x^3+C)^\frac{2}{3}$

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

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

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

Correct answer:

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

Explanation:

Rewrite $y'= \frac{x^2}{y^\frac{1}{2}}$ so that the same variables $\textup{x, dx, y, dy}$ are aligned correctly on the left and right of the equal sign.

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

$y^\frac{1}{2}\cdot \frac{dy}{dx}= x^2$

$y^\frac{1}{2}\cdot {dy}= x^2 \cdot dx$

Integrate both sides and solve for .

$\int y^\frac{1}{2}\cdot {dy}= \int x^2 \cdot dx$

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

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

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

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

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Example Question #1336 : Functions

Find the function whose slope at the point (x,y) is -4x^3+5x and passes through the point (4,5) .

Possible Answers:

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

y=181

y=0

$y= -x^4+\frac{5}{2}x^2+221$

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

Correct answer:

$y= -x^4+\frac{5}{2}x^2+221$

Explanation:

This question requires us to use differential equations. Begin as follows:

$\frac{dy}{dx}=-4x^3+5x$

dy}=-4x^3+5xdx

$y=\int -4x^3+5xdx$

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

Now, we know that y must pass through the point (4,5), so we can use this point to find c

$5= -(4)^4+\frac{5}{2}4^2+c$

5= -256+40+c

c=221

So our function is as follows:

$y= -x^4+\frac{5}{2}x^2+221$

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Example Question #31 : Solutions To Differential Equations

Solve:

$\frac{dy}{dx}=10y$

Possible Answers:

y=10e^x

y=Ce^0^.^1^x

y=Ce^1^0^x

y=Ce^x

$y=\frac{1}{10}e^x$

Correct answer:

y=Ce^1^0^x

Explanation:

Multiply and divide on both sides of the equation $\frac{dy}{dx}=10y$ .

$\frac{1}{y}dy=10 \: dx$

Integrate both sides.

$\int \frac{1}{y}dy=\int 10\: dx$

$ln\left |y \right |=10x+C$

Use base to eliminate the natural log.

$e^{ln(y)}=e^1^0^x^+^C$

y=Ce^1^0^x

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Example Question #282 : Equations

$y=e^{2x}$ .

Calculate y'(1)-y'(0).

Possible Answers:

12.78

10.2

7.39

Correct answer:

12.78

Explanation:

Remember that the derivative of $e^u = e^u \cdot u'$ .

$y'=2e^{2x}$ .

Now plug in the x=0,1 to find the corresponding values.

y'(1)=14.78

y'(0)=2.

Substitute these into the desired formula.

y'(1)-y'(0).

14.78-2=12.78

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Example Question #283 : Equations

Determine the general solution to the following differential equation:

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

Possible Answers:

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

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

$y=-2e^{-2x}+C$

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

$y=e^{-2x}+C$

Correct answer:

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

Explanation:

This is a separable differential equation, which means we can separate and , placing each on one side of the equation with its corresponding terms. There are no terms containing , so we simply place alone on one side of the equation and on the other side of the equation with any terms. We can then integrate each side with respect to the appropriate variable, which gives us an equation for that is the general solution to the differential equation:

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

$dy=(3e^{-2x})dx$

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

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

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Calculus 1 : Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #32 : Differential Equations

Example Question #37 : Differential Equations

Example Question #281 : Equations

Example Question #33 : Differential Equations

Example Question #41 : Differential Equations

Example Question #42 : Differential Equations

Example Question #1336 : Functions

Example Question #31 : Solutions To Differential Equations

Example Question #282 : Equations

Example Question #283 : Equations

All Calculus 1 Resources

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