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Calculus 1 : Equations

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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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Example Question #141 : How To Find Solutions To Differential Equations

Find the derivative of the function y=3x^2+2 .

Possible Answers:

y'(x)= 3x^2

y'(x)= 6x+2

y'(x)= 3x+2

y'(x)= 6x

Correct answer:

y'(x)= 6x

Explanation:

To find the derivative of the function, you must use the power rule

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$ .

Using this rule, the correct answer

$y'=2\cdot 3x^{2-1}$

y'(x)= 6x is obtained.

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Example Question #142 : How To Find Solutions To Differential Equations

Find the derivative of the function y=2x^2+5x^4+sin(x) .

Possible Answers:

y'(x)=4x+20x^3-sin(x)

y'(x)=2x+20x^4-cos(x)

y'(x)=4x+20x^3-cos(x)

y'(x)=4x+20x^3+cos(x)

Correct answer:

y'(x)=4x+20x^3+cos(x)

Explanation:

To take the derivative of the first two terms, you use the power rule

$\frac{\mathrm{d} }{\mathrm{d} x}(x^n)=nx^{n-1}$ .

To take the derivative of sin(x), there is the identity that,

sin'(x)=cos(x) .

Using all of the rules and applying,

$y'=2\cdot2x^{2-1}+4\cdot5x^{4-1}+cos(x)$

y'(x)=4x+20x^3+cos(x) .

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Example Question #141 : Solutions To Differential Equations

Determine the solution to the following: y'=15x^2y+12xy+y with y(0)=e^4 .

Possible Answers:

$y=\exp(5x^3+6x^2+x)$

The equation is not separable.

$y=\exp(5x^3+6x^2+x+c)$

$y=\exp(5x^3+6x^2+x+4)$

$y=\exp(5x^3+6x^2+x)+4$

Correct answer:

$y=\exp(5x^3+6x^2+x+4)$

Explanation:

Solving a differential equation tends to be quite simple if it is separable. A separable equation is one that can be rewritten in the following form

$\frac{dy}{dx}= f(x)g(y)$ .

From here we can manipulate the equation

$\frac{1}{g(y)}dy=f(x)dx$

and then you can integrate both sides.

We can rewrite the equation as

$\frac{dy}{dx}=(15x^2+12x+1)y$

and we therefore see that it is separable.

Next we manipulate it to get the the y terms on the left hand side and the x terms on the right hand side like the following

$\frac{dy}{y}=(15x^2+12x+1)dx$ .

The next step is to integrate both sides

$\int\frac{dy}{y}=\int(15x^2+12x+1)dx$ .

This gives us

$\ln y +c_1= 5x^3+6x^2+x+c_2$ ,

since the c's are just arbitrary constants we can combine them to get

$\ln y= 5x^3+6x^2+x+c$ .

Since we want to write this as an equation in terms of y we will apply e to both sides.

Therefore we get

$y=\exp(5x^3+6x^2+x+c)$ .

Since the problem gives us an initial condition we solve for c.

Therefore

$\exp(4)=\exp(5\cdot 0^3+6\cdot 0^2 +0+c)$

so c=4.

Thus the solution is

$y=\exp(5x^3+6x^2+x+4)$ .

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Example Question #144 : How To Find Solutions To Differential Equations

Solve the following differentiable equation:

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

with a condition of y(1)=0 .

Possible Answers:

y=x^2-x-1

y=x+1

y=x^2

y=x-1

y=x^2-x

Correct answer:

y=x^2-x

Explanation:

When given a first order differential equation to solve, one of the first things to check is that it is homogenous.Homogenous equations are ones in which, given

$\frac{dy}{dx}=f(x,y)$ ,

f(x,y)=f(tx,ty) given any t.

If an equation is homogenous you can do a substitution of y=xz and this new equation will be separable.

The first step is to check if the following equation is homogenous,

$\frac{dy}{dx}= \frac{x+y}{x}$ .

Next we show that the equation is homogenous

$f(x,y)=\frac{x+y}{xy}=\frac{t(x+y)}{tx}=\frac{tx+ty}{tx}=f(tx,ty)$ .

So we now replace y with xz. This gives us the following new equation:

$x\frac{dz}{dx}=\frac{x+xz}{x}= 1+z$ .

This is a separable equation which we manipulate and integrate both sides

$\int\frac{1}{x}dx=\int \frac{1}{z+1}dz$

which leads to

$\ln x = \ln(z+1)+c$ .

Thus we get the equation

x=z+1+c ,

then plugging y back in we get

$x=\frac{y}{x}+1+c$ .

So our family of equations is

y=x^2-x-cx

yet plugging in the given condition gives us that c=0.

Thus the final solution

y=x^2-x .

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Calculus 1 : Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

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

Example Question #141 : How To Find Solutions To Differential Equations

Example Question #142 : How To Find Solutions To Differential Equations

Example Question #141 : Solutions To Differential Equations

Example Question #144 : How To Find Solutions To Differential Equations

All Calculus 1 Resources

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