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Calculus 1 : Differential Equations

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Example Question #381 : Equations

Find the general solution to the following differential equation:
$\frac{\mathrm{dy} }{\mathrm{d} x}=x^3y$

Possible Answers:

$y=\ln(\frac{x^4}{4}))+C$

$y=e^{\frac{x^4}{4}}+C$

$y=Ce^{\frac{x^4}{4}}$

$y=\frac{x^4}{4}\cdot \frac{y^2}{2}+C$

Correct answer:

$y=Ce^{\frac{x^4}{4}}$

Explanation:

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Example Question #132 : Solutions To Differential Equations

f(x)=g(h(x))

$g(x)=x^{3}+x^{2}-1$

$h(x)=(x^{2}-2)$

Find f'(x) .

Possible Answers:

f'(x)=6x(x^2-2)^2+4x(x^2-2)

$f'(x)=(x^{3}+x^{2}-1)(3x+2)(2x)$

$f'(x)=2(x^{3}+x^{2}-1)$

$f'(x)= (x^{2}-2)(3x^{2}-4)$

Correct answer:

f'(x)=6x(x^2-2)^2+4x(x^2-2)

Explanation:

To find f'(x) first find f(x)=g(h(x)) .

Since,

$g(x)=x^{3}+x^{2}-1$

$h(x)=(x^{2}-2)$

then,

f(x)=(x^2-2)^3+(x^2-2)^2-1 .

We have to use the chain rule and power rule to solve this problem:

Chain Rule: $f'(x)=g'(h(x))\cdot(h'(x))$

Power Rule: $(x^n)'=nx^{n-1}$

Therefore,

$f'(x)=3(x^2-2)^{3-1}\cdot 2x+2(x^2-2)^{2-1}\cdot 2x$

f'(x)=6x(x^2-2)^2+4x(x^2-2)

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Example Question #1431 : Functions

Solve for f''(t) at t=4 .

${f}(t)= 2t^{3}-3t$

Possible Answers:

f''(4)=96

f{}''(4)=48

f{}''(4)=116

f{}''(4)= 93

Correct answer:

f{}''(4)=48

Explanation:

To solve this problem, we must first find f'(t) by deriving the original equation.

In this case, we just need to use the power rule which states,

[f(x)g(x)]'=f(x)g'(x)+g(x)f'(x) .

${f}(t)= 2t^{3}-3t$

$f{}'(t)= 6t^{2}-3$

We can then derive f'(t), again using just the power rule which states $(x^n)'=nx^{n-1}$ .

f''(t)= 12t

When we plug 4 into this final function, we arrive at the answer:

f''(4)= 12(4)=48

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Example Question #2461 : Calculus

Finding where the first derivative is positive tells us what about an equation?

Possible Answers:

Where the function is decreasing.

Where the function is concave up.

Where the function is increasing.

Where the function is concave down.

Correct answer:

Where the function is increasing.

Explanation:

Taking the first derivative gives us the function for the slope for our entire equation. Thus, finding where the slope is positive tells us where the function is increasing.

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Example Question #135 : Solutions To Differential Equations

Find the general solution to the following differential equation:

$\frac{\mathrm{dy} }{\mathrm{d} x}=x^2y+x^3y$

Possible Answers:

$y=e^{\frac{x^3}{3}+\frac{x^4}{4}}$

$y=Ce^{\frac{x^3}{3}+\frac{x^4}{4}}$

$y=Ce^{{x^3}+{x^4}$

$y=Ce^{{x^2}+x^3}$

Correct answer:

$y=Ce^{\frac{x^3}{3}+\frac{x^4}{4}}$

Explanation:

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

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

Next, we integrate both sides:

$\ln\left | y\right |=\frac{x^3}{3}+\frac{x^4}{4}+C$

The integrals were found using the following rules:

$\int \frac{dx}{x}=\ln\left | x\right |+C$ , $\int x^n dx=\frac{x^{n+1}}{n+1}+C$

To solve for y, we must exponentiate both sides of the equation:

$y=e^{\frac{x^3}{3}+\frac{x^4}{4}+C}=e^{\frac{x^3}{3}+\frac{x^4}{4}}\cdot e^C=Ce^{\frac{x^3}{3}+\frac{x^4}{4}}$

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Example Question #136 : Solutions To Differential Equations

Find the general solution for the following differential equation:

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{x^3}{e^y}$

Possible Answers:

$y=C\ln(\frac{x^4}{4})$

$y=\ln(\frac{x^4}{4})+C$

$y=e^{x^4}+C$

$y=(\frac{x^4}{4}\cdot e^y) +C$

Correct answer:

$y=C\ln(\frac{x^4}{4})$

Explanation:

To solve the separable differential equation, we must move x and dx, y and dy to their respective sides, and integrate both sides:

e^y dy=x^3 dx

$\int e^y dy=e^y+C$

$\int x^3 dx= \frac{x^4}{4}+C$

The integrations were performed using the following rules:

$\int e^x dx=e^x+C$

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

The equation then becomes

$y=\frac{x^4}{4}+C$

Note that we combine all constants of integration to a single C.

Now, we solve for y by taking the natural logarithm of both sides of the equation:

$y=\ln(\frac{x^4}{4}+C)$

Simplifying, we get

$y=C\ln(\frac{x^4}{4})$ .

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Example Question #137 : Solutions To Differential Equations

Find the solution to the following differential equation:

y'-y=0

Assume is a function of

Possible Answers:

e^t+C

Ce^t

$Ce^{-t}$

t*Ce^t

Correct answer:

Ce^t

Explanation:

Using integrating factors:

$e^{(-t)}*(y'-y)=0$

Once we use this integrating factor, we can reduce the equation to:

$\frac{\mathrm{d} }{\mathrm{d} t}[e^{-t}y]=0$

Integrating both sides:

$e^{-t}y=C$

$y=Ce^{t}$

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Example Question #131 : Solutions To Differential Equations

Determine which of the following are NOT solutions to the differential equation:

y''-y=0

Possible Answers:

y=0

y=cos(t)

y=e^t

$y=e^{-t}$

Correct answer:

y=cos(t)

Explanation:

Since you may or may not know how to solve 2nd order Differential equations, you can always plug in the answers and see if it is correct.

Let's start with y=cos(t) :

y'=-sin(t)

y''=-cos(t)

$y''-y=-cos(t)-cos(t)=-2cos(t)\neq 0$

If you try all the other function, they will solve the differential equation.

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Example Question #139 : Solutions To Differential Equations

In circuits with a resistor, the equation for voltage drop is given by:

$\frac{d}{dx}V =\frac{d}{dt}Q*R$ , where is voltage, is charge, and is resistance.

Given that Q=t^2-t+2 , and R=5 , determine the voltage difference from x:(0,1) at t=5

Possible Answers:

Correct answer:

Explanation:

This involves solving a differential equation for .

Firstly, we determine the value of $\frac{d}{dt}Q$ :

Q=t^2-t+2

$\frac{d}{dt}Q=2t-1$

At t=5

$\frac{d}{dt}Q(5)=2*5-1=9$

This means our equation is now:

$\frac{d}{dx}V =\frac{d}{dt}Q*R$

$\frac{d}{dx}V =9*5=45$

To determine the difference of from x:(0,1) , we take the definite integral:

$V=\int_{0}^{1}45 dx= 45-0=45$

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Example Question #141 : Differential Equations

Find the derivative of the function $g(x)= 5\sin(x)$

Possible Answers:

$g'(x)= -5\sin(x)$

$g'(x)= -5\cos(x)$

g'(x)= 5sin(x)

$g'(x)= 5\cos(x)$

Correct answer:

$g'(x)= 5\cos(x)$

Explanation:

The derivative of the trigonometric function $\sin(x)$ is $\cos(x)$ .

Therefore, the correct answer is

$g'(x)= 5\cos(x)$ .

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Calculus 1 : Differential Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #381 : Equations

Example Question #132 : Solutions To Differential Equations

Example Question #1431 : Functions

Example Question #2461 : Calculus

Example Question #135 : Solutions To Differential Equations

Example Question #136 : Solutions To Differential Equations

Example Question #137 : Solutions To Differential Equations

Example Question #131 : Solutions To Differential Equations

Example Question #139 : Solutions To Differential Equations

Example Question #141 : Differential Equations

All Calculus 1 Resources

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