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Calculus 1 : Differential Equations

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Example Questions

Example Question #81 : How To Find Solutions To Differential Equations

Find the general solution to the differential equation:

$\frac{\textup{dy}}{\textup{dx}}=\frac{1}{y^3\sqrt{1-x^2}}$

Possible Answers:

$y=2\sqrt{\arccos(x)}+C$

$y=\sqrt[4]{4\arccos(x)}+C$

$y=\sqrt[4]{\arccos(x)}+C$

$y=\sqrt[4]{4\arcsin(x)}+C$

Correct answer:

$y=\sqrt[4]{4\arccos(x)}+C$

Explanation:

To solve the seperable differential equation, we must put the x and dx and y and dy on the same sides:

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

Now, integrate both sides:

$\int y^3dy=\frac{y^4}{4}+C$

$\int \frac{dx}{\sqrt{1-x^2}}=\arccos(x)+C$

The integrations were performed using the following rules:

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

$\int \frac{dx}{\sqrt{1-x^2}}=\arccos(x)+C$

Finally, solve for y:

$y=\sqrt[4]{4\arccos(x)}+C$

Note that the Cs combined to make one constant of integration.

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Example Question #331 : Equations

Find the general solution to the differential equation:

$\frac{\textup{dy}}{\textup{dx}}=\frac{x^2+4}{\cos(y)}$

Possible Answers:

$y=\arccos(\frac{x^3}{3}+4x)+C$

$y=\arcsin(\frac{x^3}{3}+4x)+C$

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

$y=\ln(\sec(x))\cdot (x^2+4)+C$

Correct answer:

$y=\arcsin(\frac{x^3}{3}+4x)+C$

Explanation:

The solution for the separable differential equation can be found by first separating x and dx, y and dy:

$\cos(y)dy=(x^2+4)dx$

Now, integrate both sides:

$\int \cos(y)dy=\sin(y)+C$ , $\int (x^2+4)dx=\frac{x^3}{3}+4x+C$

The following rules were used for integration:

$\int \cos(x)dx=\sin(x)+C$ , $\int x^ndx=\frac{x^{n+1}}{n+1}+C$

Finally, solve for y:

$y=\arcsin(\frac{x^3}{3}+4x)+C$

Note that the Cs combined to make one constant of integration.

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Example Question #81 : How To Find Solutions To Differential Equations

Find the derivative of the function

$f(x)=\frac{sin(x)cos(x)}{x^2}$ .

Possible Answers:

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

$\frac{x(cos^2(x)-sin^2(x))-2sin(x)cos(x)}{x^5}$

$\frac{(cos^2(x)-sin^2(x))-2sin(x)cos(x)}{x^4}$

$\frac{x(cos^2(x)-sin^2(x))+2sin(x)cos(x)}{x^5}$

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

Correct answer:

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

Explanation:

To find this derivative, use of both the product rule and quotient rule for derivatives will be necessary. The latter states:

$\frac{d}{dx}\frac{u}{v}=\frac{vdu-udv}{v^2}$

For our function

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

u=sin(x)cos(x);v=x^2

The v=x^2 derivative is simply dv=2xdx

For the u=sin(x)cos(x) derivative, the product rule will be useful:

$\frac{d}{dx}st=sdt+tds$

Where s=sin(x);t=cos(x)

And the derivative is:

du=(-sin^2(x)+cos^2(x))dx

Putting all of this together, the derivative of f(x) is:

$\frac{d}{dx}f(x)=\frac{x^2(cos^2(x)-sin^2(x))-2xsin(x)cos(x)}{x^4}$

$\frac{d}{dx}f(x)=\frac{x(cos^2(x)-sin^2(x))-2sin(x)cos(x)}{x^3}$

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Example Question #333 : Equations

Find the general solution to the given differential equation:

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

Possible Answers:

$y=\arccos \left(\frac{3x^4}{4}\right)+C$

$y=\arcsin\left(\frac{3x^4}{4}\right)+C$

$y=-\arcsin\left(\frac{3x^4}{4}\right)+C$

$y=\frac{3x^4}{4}\cdot \ln\left | \sec(x)+ \tan(x)\right |+C$

Correct answer:

$y=\arcsin\left(\frac{3x^4}{4}\right)+C$

Explanation:

To solve the differential equation, we must move the x and y terms with dx and dy, respectively:

$\cos(y)dy=3x^3 dx$

Now we can integrate:

$\int \cos(y)dy=\sin(y)=\int 3x^3dx=\frac{3x^4}{4}+C$

using the following rules:

$\int x^n dx= \frac{x^{n+1}}{n+1}+C$ , $\int cos(x)dx=\sin(x)+C$

To finish, write the equation in terms of y alone:

$y=\arcsin\left(\frac{3x^4}{4}\right)+C$

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Example Question #334 : Equations

Find the derivative of

$\sin(x)^2+1{}{}$ .

Possible Answers:

${} 2\sin(x)^2$

${} \cos(x)^2$

${} 2\sin(x)\cdot\cos(x)$

${} \sin(x)^2\cdot\cos(x)$

Correct answer:

${} 2\sin(x)\cdot\cos(x)$

Explanation:

This can be easily separated into two derivatives added together:

${} D_x[\sin(x)^2]+D_x[1]$

The second function is easy: the derivative of any constant is 0. But for the first, we must use the chain rule.

Recall:

${} (f\circ g)'=(f'\circ g)\cdot g'$

Our outside function is x^2 and our inside function is sin(x) .

So the chain rule tells us we must take the derivative of x^2 and plug sin(x) into that function.

The derivative of x^2 is , so we have

${} 2\sin(x)\cdot \sin(x)'$ .

Now all we have to do is find the derivative of sin(x) , which we know is cos(x) .

So our final answer is

${} 2\sin(x)\cdot \cos(x)$ .

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Example Question #335 : Equations

Find the solution to the differential equation.

$\frac{dy}{dt}= 3t^2-4t$

Possible Answers:

y= 6t-4

y= t^3-2t^2+C

y= 3t^3-4t^2+C

y= C

Correct answer:

y= t^3-2t^2+C

Explanation:

$\frac{dy}{dt}= 3t^2-4t$

$\int \frac{dy}{dt}dt= \int (3t^2-4t)dt$

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.

In our case,

$y= \int (3t^2-4t)dt= \int (3t^2)dt-\int (4t)dt=t^3-2t^2+C$ , where is a constant.

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Example Question #81 : How To Find Solutions To Differential Equations

Which type of function solves the differential equation given.

$\frac{dy}{dt}=\frac{2y}{t}$

Possible Answers:

Polynomial

Rational

Logarithmic

Exponential

Correct answer:

Polynomial

Explanation:

Without doing very much, we can see that the solution to this differential equation will be such that the derivative is equal to the original function divided by .

One such type of function is a polynomial, where the derivative decreases the order of the function by , which is equivalent to dividing by .

We can also show this through integration.

$\frac{dy}{dt}=\frac{2y}{t}$

First separate the variables.

$\frac{dy}{y}=\frac{2}{t}dt$

Now take the integral remembering the rule for natural logs.

$\int\frac{1}{x}=ln(x)$

Thus we get,

$\\ \int \frac{dy}{y}=\int \frac{2}{t}dt \\ \\ln(y)=2ln(t)$

From here recall the properties of natural logs such that cln(x)=ln(x^c) .

Therefore we can rewrite our function and exponentiate it to solve for y.

$\\ e^{ln(y)}=e^{ln(t^2)} \\y=t^2$ .

From here, we can see that it is a polynomial.

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Example Question #341 : Equations

Find the general solution to the differential equation:

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

Possible Answers:

$y=\arctan (e^{2x})+C$

$y=\frac{e^{2x}\cdot \ln\left | \tan(y)\right |}{2}+C$

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

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

Correct answer:

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

Explanation:

To find the general solution to the seperable differential equation, we must seperate the x and dx, y and dy to seperate sides:

$\sec^2(y)dy=e^{2x}dx$

Now, integrate both sides (note that the two constants of integration combine to make one C):

$\tan(y)=\frac{1}{2}e^{2x}+C$

The following rules were used for integration:

$\int \sec^2(x)dx=\tan(x)+C$ , $\int e^udx=\frac{e^u}{\frac{du}{dx}}+C$

Finally, solve for y:

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

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Example Question #83 : How To Find Solutions To Differential Equations

Where will $\cos(2x)$ have a local minimum or maximum between and $\pi$ ?

Possible Answers:

$\frac{\pi}{}4$

, $\frac{\pi}{}2$

. $\frac{\pi}{}4$ , $\frac{\pi}{}2$ , $\frac{3\pi}{}4$

$\frac{\pi}{}4$ , $\frac{3\pi}{}4$

Correct answer:

, $\frac{\pi}{}2$

Explanation:

We must find the maximums and minimumss by setting the derivative equal to zero.

The derivative of $\cos(2x)$ is found using the chain rule,

$\frac{d}{dx}f(g(x))=f'(g(x))g'(x)$

and the trigonometric rule of derivatives for cosine,

$\frac{d}{dx}cos(x)=-sin(x)$ .

Applying the above rules we find the derivative to be,

$-2\sin(2x)$ . $\sin(x)=0$ when $x=0, \pi$ , so in this case, we must divide each angle by two since we are dealing with $\sin(2x)$ , so $x=0, \frac{\pi}{2}$ .

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Example Question #90 : How To Find Solutions To Differential Equations

Which one of the following functions is continuous and has continuous derivative at 2?

Possible Answers:

$f(x)=\begin{Bmatrix} 4x^2&x\geq 2\\ 8x & x<2 \end{Bmatrix}$

$f(x)=\begin{Bmatrix} 2x^2+4&x\geq 2\\ x-4 & x<2 \end{Bmatrix}$

$f(x)=\begin{Bmatrix} 2x^2+2&x\geq 2\\ 8x-8 & x<2 \end{Bmatrix}$

$f(x)=\begin{Bmatrix} 2x^2+4&x\geq 2\\ 8x-4 & x<2 \end{Bmatrix}$

$f(x)=\begin{Bmatrix} x^2+4&x\geq 2\\ 8x-4 & x<2 \end{Bmatrix}$

Correct answer:

$f(x)=\begin{Bmatrix} 2x^2+4&x\geq 2\\ 8x-4 & x<2 \end{Bmatrix}$

Explanation:

In order for the graph to be continuous, both of the parts of the piecewise function must be equal.

For the derivative to be equal, one must take the derivatives at each part and see if this new piecwise function is equal at 2 as well.

Only

$f(x)=\begin{Bmatrix} 2x^2+4&x\geq 2\\ 8x-4 & x<2 \end{Bmatrix}$

fufills both these requirements.

Remember to use the power rule to find the derivatives of the piecewise function, $\frac{d}{dx}x^n=nx^{n-1}$ .

$\\ f(2)=2(2)^2+4=12 \\f(2)=8(2)-4=16-4=12\\f'(2)=4(2)=8\\f'(2)=8$

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Calculus 1 : Differential Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #81 : How To Find Solutions To Differential Equations

Example Question #331 : Equations

Example Question #81 : How To Find Solutions To Differential Equations

Example Question #333 : Equations

Example Question #334 : Equations

Example Question #335 : Equations

Example Question #81 : How To Find Solutions To Differential Equations

Example Question #341 : Equations

Example Question #83 : How To Find Solutions To Differential Equations

Example Question #90 : How To Find Solutions To Differential Equations

All Calculus 1 Resources

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