How to find the solution of a differential equation

All Calculus 1 Resources

Find the derivative of (5+3x)5.

Possible Answers:

5x(5+3x)^4

15(5+3x)^4

5(5+3x)^4x

15x(5+3x)^4

5(5+3x)^4

Correct answer: 15(5+3x)^4

Explanation:

We'll solve this using the chain rule.

Dx[(5+3x)5]

=5(5+3x)4 * Dx[5+3x]

=5(5+3x)4(3)

=15(5+3x)4

Find Dx[sin(7x)].

Possible Answers:

7sin(7x)

-7sin(7x)

7sin(7x)cos(7x)

7cos(7x)

-7cos(7x)

Correct answer: 7cos(7x)

Explanation:

First, remember that Dx[sin(x)]=cos(x). Now we can solve the problem using the Chain Rule.

Dx[sin(7x)]

=cos(7x)*Dx[7x]

=cos(7x)*(7)

=7cos(7x)

Calculate fxxyz if f(x,y,z)=sin(4x+yz).

Possible Answers:

-16sin(4x+yz)

4sin(4x+yz)

-16cos(4x+yz) +16yzsin(4x+yz)

arctan(4x+yz)

cos(4x+yz)

Correct answer: -16cos(4x+yz) +16yzsin(4x+yz)

Explanation:

We can calculate this answer in steps.  We start with differentiating in terms of the left most variable in "xxyz".  So here we start by taking the derivative with respect to x. 

First, fx= 4cos(4x+yz)

Then, fxx= -16sin(4x+yz)

fxxy= -16zcos(4x+yz)

Finally, fxxyz= -16cos(4x+yz) + 16yzsin(4x+yz) 

Integrate 

Correct answer:

Integrate :

Correct answer:

Find the general solution, , to the differential equation

 .

Correct answer:

Explanation:

We can use separation of variables to solve this problem since all of the "y-terms" are on one side and all of the "x-terms" are on the other side.  The equation can be written as .  

Integrating both sides gives us . 

Consider  ; by multiplying by  both the left and the right hand sides can be swiftly integrated as

 

where .  So, for example,  can be rewritten as: 

. We will use this trick on another simple case with an exact integral.

Correct answer:

Explanation:

As described in the problem,  we are given

.

We can multiply both sides by :

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

This yields:

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

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:

,

which we learned to solve by the trigonometric substitution, yielding:

Clearly  and the fact that  again gives us  so

What are all the functions  such that

?

Possible Answers:

 for arbitrary constants k and C

 for arbitrary constants k and C

 for arbitrary constants k and C

 for arbitrary constants k and C

 for arbitrary constants k and C

Correct answer:

 for arbitrary constants k and C

Explanation:

Integrating once, we get:

Integrating a second time gives:

We integrate the first term by parts using  to get:

Canceling the x's we get:

Defining  gives the above form.

The Fibonacci numbers are defined as 

and are intimately tied to the golden ratios , which solve the very similar equation

.

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

Find the Fibonacci function defined by:

whose derivatives at 0 are therefore the Fibonacci numbers.

Correct answer:

Explanation:

To solve , we ignore  of the derivatives to get simply:

This can be solved by assuming an exponential function , which turns this expression into

,

which is solved by  . Our general solution must take the form:

Plugging in our initial conditions  and , we get:

Hence the answer is:

Find the particular solution given . 

Correct answer:

Explanation:

The first thing we must do is rewrite the equation:

We can then find the integrals: 

The integrals as as follows:

we're left with 

We then plug in the initial condition and solve for 

The particular solution is then:

All Calculus 1 Resources

Which function is a solution to the differential equation xy '+ 2y 0?

Summary: The solution to the differential equation xy' + 2y = 0 is x−2 .

How many solution can we find for a differential equation?

As we have seen so far, a differential equation typically has an infinite number of solutions. Such a solution is called a general solution . A corresponding initial value problem will give rise to just one solution. Such a solution in which there are no unknown constants remaining is called a specific solution .

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