How to find solutions to differential equations - AP Calculus AB

Card 0 of 2016

Question

Find the particular solution for the following initial value problem:

$x\frac{dy}{dx}-y\sqrt{x}=0$

Answer

To find the particular solution, we start by finding the general solution. First we rearrange the differential equation such that is on one side with any terms and is on the other side with any terms. We can then integrate each side with respect to the appropriate variable and solve for to find the general solution for the differential equation. Finally, we plug in the given initial condition to determine the value of the constant, which gives us the particular solution:

$x\frac{dy}{dx}-y\sqrt{x}=0$

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

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

$\frac{1}{y}dy=\frac{1}{\sqrt{x}}dx$

$\int \frac{1}{y}dy=\int \frac{1}{\sqrt{x}}dx$

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

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

$y=e^{2\sqrt{x}+C}=e^Ce^{2\sqrt{x}}=Ce^{2\sqrt{x}}$

$y(0)=1\rightarrow Ce^{2\sqrt{0}}=1\rightarrow C=1$

$y=e^{2\sqrt{x}}$

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Question

Find the particular solution for the following differential equation:

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

Answer

To find the particular solution, we start by finding the general solution. First we rearrange the differential equation such that is on one side with any terms and is on the other side with any terms. We can then integrate each side with respect to the appropriate variable and solve for to find the general solution for the differential equation. Finally, we plug in the and values of the given point to determine the value of the constant, which gives us the particular solution:

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

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

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

$\ln(y)=-x^{-1}+C$

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

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

$y(1)=-1\rightarrow Ce^{-\frac{1}{1}}=-1\rightarrow C=-e$

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

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Question

Find for the following equation:

$f(x)=\ln(x)+7x^2-19x$

Answer

First, find the derivative. Then, evaluate at x=3.

For this function we will use the Power Rule to find the derivative.

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$

Also remember that the derivative of $\ln(x)$ is $\frac{1}{x}$ .

Therefore we get,

$f(x)=\ln(x)+7x^2-19x$

$f'(x)=\frac{1}{x}+14x-19$

$f'(3)=\frac{1}{3}+14(3)-19=23\frac{1}{3}$

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Question

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

Answer

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

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Question

Find Dx\[sin(7x)\].

Answer

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)

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Question

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

Answer

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)

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Question

Integrate $\small \int_{0}^{\pi } \cos x \hspace 2 dx$

Answer

$\frac{\mathrm{d} }{\mathrm{d} x}\sin x=cosx$

thus:

$\small \small \int_{0}^{\pi } \cos x \hspace 2 dx = \sin x \left]^\pi_0$
$\small \sin \pi - \sin 0 = 0 - 0 = 0$

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Question

Integrate :

$\small \int_{-\frac{\pi}{4}}^{0} \sec x \tan x \hspace 2 dx$

Answer

$\frac{\mathrm{d} }{\mathrm{d} x}\sec x=\sec x\tan x$

thus:
$\small \small \int_{-\frac{\pi}{4}}^{0} \sec x \tan x \hspace 2 dx = \sec x \left]^0_{-\frac{\pi}{4} }$
$\small = \sec 0 - \sec(-\frac{\pi}{4}) = 1 - \sqrt{2}$

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Question

Find the general solution, , to the differential equation

$\frac{dy}{dx}=30-2x$ .

Answer

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 $\frac{dy}{dx}=30-2x\Rightarrow dy=(30-2x)dx$ .

Integrating both sides gives us $\int dy = \int (30-2x) dx\Rightarrow y(x) = 30x-x^2+C$ .

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Question

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

$\frac{1}{2}(y')^2 = F(y) + C$

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

$\frac{1}{2} [y'(x)]^2 = \frac{1}{3} [y(x)]^3 + C$ . We will use this trick on another simple case with an exact integral.

Use the technique above to find such that with and .

Hint: Once you use the above to simplify the expression to the form , you can solve it by moving into the denominator:

$\int dy/g(y) = \int dx$

Answer

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:

$\frac{1}{2}[(y')^2]' = [y^2]' + \frac{1}{2} [y^4]'$

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:

$\frac{dy}{dx} = \pm (1 + y^2)$

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:

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

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

$\tan^{-1} y = x + C$

Clearly and the fact that again gives us so

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Question

What are all the functions such that

?

Answer

Integrating once, we get:

$y'(x) = \int \frac{dx}{x} + C_1 = \ln x + C_1$

Integrating a second time gives:

$y(x) = \int \ln x ~ dx + C_1 ~ x + C$

We integrate the first term by parts using to get:

$y(x) = x ~ \ln x ~-~ \int x ~ \frac{dx}{x} ~+~ C_1 ~ x ~+~ C$

Canceling the x's we get:

$y(x) = x ~ \ln x ~+~ \left( C_1 - 1 \right ) x ~+~ C$

Defining gives the above form.

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Question

The Fibonacci numbers are defined as

$F_0 = 0, F_1 = 1, F_n = F_{n-1} + F_{n-2}$

and are intimately tied to the golden ratios $\phi_\pm = (1 \pm \sqrt{5})/2$ , which solve the very similar equation

$\phi^n_\pm = \phi^{n - 1}_\pm + \phi^{n-2}_\pm$ .

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

$y^{(0)}(x) = y(x), y^{(n)}(x) = \frac{d}{dx} y^{(n - 1)}(x)$

Find the Fibonacci function defined by:

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

$f^{(0)}(0) = 0, f^{(1)}(0) = 1$

whose derivatives at 0 are therefore the Fibonacci numbers.

Answer

To solve $f^{(n)} = f^{(n-1)} + f^{(n-2)}$ , we ignore of the derivatives to get simply:

This can be solved by assuming an exponential function $y = C e^{k x}$ , which turns this expression into

,

which is solved by $k = \phi_\pm$ . Our general solution must take the form:

$f(x) = C_+ ~ e^{x \phi_+} + C_- ~ e^{x \phi_-}$

Plugging in our initial conditions and , we get:

$C_+ \ \phi_+ ~+~ C_- \ \phi_- = 1$

$C_+ = 1/(\phi_+ - \phi_-) = 1/(\frac{1+\sqrt 5}{2} - \frac{1-\sqrt 5}{2}) = 1/\sqrt{5}$

Hence the answer is:

$f(x) = (e^{x \phi_+} - e^{x \phi_-}) / \sqrt{5}$

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Question

Find of the following equation:

Answer

First take the derivative and then solve when x=2.

To find the derivative use the power rule which states when,

the derivative is $f'(x)=nx^{n-1}$ .

Therefore the derivative of our function is:

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Question

Find for the following equation:

Answer

To find the derivative of this function we will need to use the product rule which states to multiply the first function by the derivative of the second function and add that to the product of the second function and the derivative of the first function. In other words,

$f'(x)=h(x)\cdot g'(x)+g(x)\cdot h'(x)$

To do this we will let,

and

and

Now we can find the derivative by plugging in these equations as follows.

Now plug in x=1 and solve.

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Question

Find the solution to the following equation at

Answer

To solve, we must first find the derivative and then solve when x=-2.

To find the derivative of the function we will use the Power Rule:

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$

Therefore,

Now to solve for -2 we plug it into our x value.

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Question

Find the particular solution given .

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{9x^2}{y^2-4y}$

Answer

The first thing we must do is rewrite the equation:

We can then find the integrals:

$\int (y^2-4y)dy= \int (9x^2)dx$

The integrals as as follows:

$\int (y^2-4y)dy= \frac{1}{3}y^3-2y^2$

$\int (9x^2)dx=3x^3$

we're left with

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

We then plug in the initial condition and solve for

$\frac{1}{3}(3)^3-2(3)^2=3(1)^3+c$

The particular solution is then:

$\frac{1}{3}y^3-2y^2=3x^3-12$

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Question

Find the particular solution given $y ( \frac{1}{2} )= -1$ .

$\frac{\mathrm{d}y }{\mathrm{d} x}=2xy^2$

Answer

The first thing we must do is rewrite the equation:

$\left(\frac{1}{y^2}\right)dy=(2x) dx$

We can then find the integrals:

$\int \left(\frac{1}{y^2}\right)dy=\int (2x) dx$

_Th_e integrals are as follows:

$\int \left(\frac{1}{y^2}\right)dy= \int y^{-2} = -1y^{-1}=-\frac{1}{y}$

$\int (2x) dx=x^2$

We're left with:

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

We then plug in the initial condition and solve for

$-\frac{1}{(-1)}=\left(\frac{1}{2}\right)^2+c$

$1=\left(\frac{1}{4}\right)+c$

$c=\frac{3}{4}$

The particular solution is then:

$-\frac{1}{y}=x^2+\frac{3}{4}$

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Question

Find the particular solution given $y\left(-\frac{1}{2}\right)=-\frac{1}{4}$ .

$\frac{\mathrm{d} y}{\mathrm{d} x}=6x^2y^2$

Answer

The first thing we must do is rewrite the equation:

$\left(\frac{1}{y^2}\right)dy=(6x^2)dx$

We can then find the integrals:

$\int \left(\frac{1}{y^2}\right)dy=\int (6x^2)dx$

The integrals are as follows:

$\int \left(\frac{1}{y^2}\right)dy=-\frac{1}{y}$

$\int (6x^2)dx=2x^3$

We're left with:

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

We then plug in the initial condition and solve for

$-\frac{1}{(-\frac{1}{4})}=2\left(-\frac{1}{2}\right)^3+c$

$4=-\frac{1}{4}+c$

$c=\frac{17}{4}$

The particular solution is then:

$-\frac{1}{y}=2x^3+\frac{17}{4}$

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Question

Find the particular solution given $y(4)=\frac{256}{9}$ .

$\frac{\mathrm{d} y}{\mathrm{d} x}=\sqrt{xy}$

Answer

Remember: $\sqrt{xy}=(\sqrt{x} ) (\sqrt{y})$

The first thing we must do is rewrite the equation:

$\left(\frac{1}{\sqrt{y}}\right)dy=(\sqrt{x})dx$

We can then find the integrals:

$\int \left(\frac{1}{\sqrt{y}}\right)dy=\int (\sqrt{x})dx$

The integrals are as follows:

$\int \left(\frac{1}{\sqrt{y}}\right)dy= \int y^{-1/2}dy=\frac{y^{1/2}}{\frac{1}{2}}=2y^{1/2}=2\sqrt{y}$

$\int (\sqrt{x})dx=\frac{2}{3}(x^{\frac{3}{2}})$

We're left with:

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

We then plug in the initial condition and solve for

$2\sqrt{\frac{256}{9}}= \frac{2}{3}(4)^{\frac{3}{2}}+c$

$\frac{32}{3}=\frac{16}{3}+c$

$c=\frac{16}{3}$

The particular solution is then:

$2\sqrt{y}=\frac{2}{3}(x)^{\frac{3}{2}}+\frac{16}{3}$

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Question

Find the particular solution given .

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{5x^4}{y^2+4y^3}$

Answer

The first thing we must do is rewrite the equation:

We can then find the integrals:

$\int (y^2+4y^3)dy=\int (5x^4)dx$

The integrals are as follows:

$\int (y^2+4y^3)dy=\frac{1}{3}y^3+y^4$

$\int (5x^4)dx=x^5$

We're left with

$\frac{1}{3}y^3+y^4=x^5+c$

We plug in the initial condition and solve for

$\frac{1}{3}(0)^3+(0)^4=(1)^5+c$

The particular solution is then:

$\frac{1}{3}y^3+y^4=x^5-1$

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