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Calculus 1 : Differential Equations

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Example Question #61 : How To Find Solutions To Differential Equations

Find the solution to the differential equation

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

with initial value $y(\pi/2)=2$ .

Possible Answers:

$y(x)=e^{\sin x}$

$y(x)=\frac{2}{e}e^{\sin x}$

$y(x) = 2e^{\sin x}$

$y(x) = 2e\cdot e^{\sin x}$

Correct answer:

$y(x)=\frac{2}{e}e^{\sin x}$

Explanation:

We can solve this by method of separating variables:

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

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

So we integrate to get

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

$\iff \ln y = \sin x+c$

$\iff e^{\ln y}=y=e^{\sin x+c}=e^{\sin x}e^c$

$\Rightarrow y(x)=e^{\sin x}e^c$

So now we plug in the initial value $y(\pi/2)=2$ to get the specific solution:

$\small y(\pi/2)=2=e^{\sin \pi/2}e^c$

$\small \iff 2 = e^1e^c=e^{1+c}$

So we solve for $\small c$ to get $\small \small c=\ln 2-1$ . We plug this $\small c$ into $\small y(x)$ to get

$\small y(x)=e^{\sin x}e^{\ln 2-1}$

which becomes

$\small y(x)=\frac{2}{e}e^{\sin x}$

after simplifying.

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Example Question #62 : How To Find Solutions To Differential Equations

Find the general solution to the differential equation

$\small \frac{dy}{dt}=yt^3$ .

Possible Answers:

$\small \small y=e^{3t^2}e^c$

$\small \small \small y=e^{\frac{t^4}{4}}+e^c$

$\small \small \small y=e^{3t^2}+e^c$

$\small y=e^{\frac{t^4}{4}}e^c$

Correct answer:

$\small y=e^{\frac{t^4}{4}}e^c$

Explanation:

We can use separation of variables to solve the differential equation

$\small \frac{dy}{dt}=yt^3$

We have

$\small \frac{1}{y}dy=t^3dt$

$\small \small \int\frac{1}{y}dy=\int t^3dt$

$\small \small \implies \ln y=\frac{t^4}{4}+c$

$\small \small \small \iff e^{\ln y}=y=e^{\frac{t^4}{4}+c}=e^{\frac{t^4}{4}}e^{c}$

So the general solution is $\small y=e^{\frac{t^4}{4}}e^c$ , or $\small \small y=Ke^{\frac{t^4}{4}}$ for $\small K=e^c$ .

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Example Question #63 : How To Find Solutions To Differential Equations

What is the solution to the differential equation

$\frac{df}{dx}=sec^3(x)tan(x)?$

Possible Answers:

f(x)=3sec^3(x)tan^2(x)+sec^5(x)+c

$f(x)=\frac{sec^3(x)}{3}+c$

$f(x)=\frac{sec^4(x)tan^2(x)}{8}+c$

$f(x)=\frac{sec^4(x)}{4}+c$

Correct answer:

$f(x)=\frac{sec^3(x)}{3}+c$

Explanation:

First, multiply both sides of the equation by dx to find

df=sec^3(x)tan(x)dx

Second, integrate both sides of the equation

$\int df=\int sec^3(x)tan(x)dx$

To integrate the right side, we can use u-substitution. Let u=sec(x). Then du=sec(x)tan(x)dx. Therefore we can rewrite the integral in terms of u as:

$\int df=\int u^2du$

Next take the antiderivative of both sides

$f=\frac{u^3}{3}+c$

Substitute sec(x) for u to find the final solution in terms of x.

$f(x)=\frac{sec^3(x)}{3}+c$

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Example Question #64 : How To Find Solutions To Differential Equations

Find the solution to the separable differential equation.

$\frac{dy}{dx}=xy+2y$

Possible Answers:

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

$y=ln\left(\frac{x^2}{2}+x\right)+c$

$y=ce^{\frac{x^2}{2}+2x}$

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

Correct answer:

$y=ce^{\frac{x^2}{2}+2x}$

Explanation:

To solve this separable differential equation, we first need to rewrite it so that the left side is expressed entirely in terms of y and the right side in terms of x.

First factor the right side of the equation:

$\frac{dy}{dx}=xy+2y \rightarrow \ \frac{dy}{dx}=y(x+2)$

Next, divide both sides of the equation by y and multiply both sides of the equation by dx, which results in:

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

Integrate both sides of the equation:

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

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

Finally, exponentiate both sides of the equation to solve for y:

$e^{ln(y)}=e^{\frac{x^2}{2}+2x+c}$

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

Finally, e^c is a constant, so just write it as 'c'.

$y=ce^{\frac{x^2}{2}+2x}$

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Example Question #65 : How To Find Solutions To Differential Equations

Find the derivative of the following function:

$y=(3x+7)(4x-3)^{3}$

Possible Answers:

$\frac{dy}{dx}=(12x+28)(4x-3)^{2}+3(4x-3)^3$

$\frac{dy}{dx}=(9x+21)(4x-3)^{2}+3(4x-3)^{3}$

$\frac{dy}{dx}=(12x-28)(4x-3)^2+3(4x-3)^3$

$\frac{dy}{dx}=(36x+84)(4x-3)^{2}+3(4x-3)^{3}$

$\frac{dy}{dx}=(21x+28)+3(4x-3)^{3}$

Correct answer:

$\frac{dy}{dx}=(36x+84)(4x-3)^{2}+3(4x-3)^{3}$

Explanation:

Computation of the derivative requires use of the Product Rule and the Chain Rule.

$y=(3x+7)(4x-3)^{3}$

$\frac{dy}{dx}=\frac{d}{dx}[(3x+7)(4x-3)^3]$

A good way to remember the Product Rule is by memorizing this saying: "First Times the Derivative of the Second, Plus the Second Times the Derivative of the First." Or if that doesn't help then you can just write out the formula:

For:

y = f(x)*g(x) Where f(x) and g(x) are differentiable functions

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

As you can see, the "saying" from above matches the formula.

In this case:

f(x)=3x+7 , g(x)=(4x-3)^3

Applying the Product Rule:

$\frac{dy}{dx}=(3x+7)\frac{d}{dx}[(4x-3)^3]+(4x-3)^3\frac{d}{dx}[(3x+7)]$

To compute the derivatives of (4x-3)^3 and (3x+7) , simply apply the chain rule:

For:

y=(f(x))^a

$\frac{dy}{dx}=a*(f(x))^{a-1}*f'(x)$

Where a is an integer and f(x) is a differentiable function.

Applying the Chain Rule:

$\frac{dy}{dx}=(3x+7)[3(4x-3)^2(4)]+(4x-3)^3(3)$

Simplify the expression to match one of the answer choices:

$\frac{dy}{dx}=(36x+84)(4x-3)^2+3(4x-3)^3$

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Example Question #66 : How To Find Solutions To Differential Equations

Find the derivative with respect to for the following function:

$y=\frac{3x^2-5}{(2x^2-7)^3}$

Possible Answers:

$\frac{dy}{dx}=\frac{-24x^3+18x}{(2x^2-7)^4}$

$\frac{12x^3-36x^2+12x}{(2x^2-7)^5}$

$\frac{12x^3-78x^2-60x}{(2x^2-7)^3}$

$\frac{12x^3-36x^2-102x}{(2x^2-7)^5}$

$\frac{24x^3+18x}{(2x-7)^4}$

Correct answer:

$\frac{dy}{dx}=\frac{-24x^3+18x}{(2x^2-7)^4}$

Explanation:

Computation of the derivative requires use of the Quotient Rule and Chain Rule.

$y=\frac{3x^2-5}{(2x^2-7)^3}$

$\frac{dy}{dx}=\frac{d}{dx}\left[\frac{3x^2-5}{(2x^2-7)^3}\right]$

A good way to memorize the Quotient Rule is by memorizing this saying: "Bottom Times the Derivative of the Top, Minus the Top Times the Derivative of the Bottom, All Over the Bottom Squared." Or if that doesn't help then you can just write out the formula:

For:

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

Where f(x) and g(x) are differentiable functions where g(x) Does Not Equal 0!

$\frac{dy}{dx}=\frac{g(x)*f'(x)-f(x)*g'(x)}{(g(x))^2}$

As you can see, the "saying" from above matches the formula.

In this case:

f(x)=3x^2-5 , g(x)=(2x^2-7)^3

Applying the Quotient Rule:

$\frac{dy}{dx}=\frac{(2x^2-7)^3\frac{d}{dx}[3x^2-5]-(3x^2-5)\frac{d}{dx}[(2x^2-7)^3]}{((2x^2-7)^3)^2}$

To compute the derivatives of 3x^2-5 and (2x^2-7)^3 , simply apply the chain rule:

For:

y=(f(x))^a

$\frac{dy}{dx}=a*(f(x))^{a-1}*f'(x)$

Where a is an integer and f(x) is a differentiable function.

Applying the Chain Rule:

$\frac{dy}{dx}=\frac{(2x^2-7)^3(6x)-(3x^2-5)[3(2x^2-7)^2(4x)]}{(2x^2-7)^6}$

We can simplify this equation further by factoring out a common (2x^2-7)^2 from the numerator.

$\frac{dy}{dx}=\frac{(2x^2-7)^2[6x(2x^2-7)-(3x^2-5)(12x)]}{(2x^2-7)^6}$

You will notice that the can cancel out with some terms from the denominator resulting in:

$\frac{dy}{dx}=\frac{(6x)(2x^2-7)-(3x^2-5)(12x)}{(2x^2-7)^4}$

Distributing the terms out in the numerator and simplifying the expression results in the final answer, which matches one of the answer choices:

$\frac{dy}{dx}=\frac{-24x^3+18x}{(2x^2-7)^4}$

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Example Question #67 : How To Find Solutions To Differential Equations

Find the derivative of the following function.

$f(x)=(3x^3+5x^2-7)^{^{10}}$

Possible Answers:

$10(3x^3+5x^2-7)^{9}$

$(9x^2+10x)(3x^3+5x^2-7)^{9}$

$(9x^2+10x)(300x^3+50x^2-70)^{9}$

$(90x^2-100x)(3x^3+5x^2-7)^{9}$

$(90x^2+100x)(3x^3+5x^2-7)^{9}$

Correct answer:

$(90x^2+100x)(3x^3+5x^2-7)^{9}$

Explanation:

$D_x[(3x^3+5x^2-7)^{10}]$ Finding the derivative requires use of the chain rule.

$D_x[(f(x))^a]=a*f(x)^{a-1}*f'(x)$ Where f(x) is a differentiable function and a is an integer.

From the problem statement:

f(x)=3x^3+5x^2-7

Computing the derivative:

f'(x)=9x^2+10x

Applying the Chain Rule:

$D_x[(3x^3+5x^2-7)^{10}]=10*(3x^3+5x^2-7)^{9}*(9x^2+10x)$

Simplify to match the answer choice:

$(90x^2+100x)(3x^3+5x^2-7)^{9}$

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Example Question #68 : How To Find Solutions To Differential Equations

For:

y=sin(x^2+2)cos(x-3)

Find $\frac{dy}{dx}$ :

Possible Answers:

2xsin(x-3)cos(x^2+2)-cos(x-3)sin(x^2+2)

2xcos(x-3)cos(x^2+2)-sin(x-3)sin(x^2+2)

2xsin(x-3)sin(x^2+2)-cos(x-3)cos(x^2+2)

2xcos(x^2+2)-sin(x^2+2)

2xcos(x+3)cos(x^2+2)-sin(x+3)sin(x^2+2)

Correct answer:

2xcos(x-3)cos(x^2+2)-sin(x-3)sin(x^2+2)

Explanation:

Computation of the derivative requires use of the Product Rule and Chain Rule.

y=sin(x^2+2)cos(x-3)

$\frac{dy}{dx}=\frac{d}{dx}[sin(x^2+2)cos(x-3)]$

For:

y=f(x)*g(x) Where f(x) and g(x) are differentiable functions

As you can see, the "saying" from above matches the formula.

In this case:

f(x)=sin(x^2+2) , g(x)=cos(x-3)

Applying the Product Rule:

$\frac{dy}{dx}=sin(x^2+2)\frac{d}{dx}[cos(x-3)]+cos(x-3)\frac{d}{dx}[sin(x^2+2)]$

To compute the derivatives of cos(x-3) and sin(x^2+2) , simply apply the chain rule:

For:

y=sin(u)

$\frac{dy}{dx}=cos(u)\frac{du}{dx}$ , Where u is a differentiable function

For:

y=cos(u)

$\frac{dy}{dx}=-sin(u)\frac{du}{dx}$ , Where u is a differentiable function

Applying the Chain Rule:

$\frac{dy}{dx}=sin(x^2+2)(-sin(x-3)(1))+cos(x-3)(cos(x^2+2)(2x))$

Simplify the expression to match one of the answer choices:

$\frac{dy}{dx}=2xcos(x-3)cos(x^2+2)-sin(x-3)sin(x^2+2)$

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Example Question #69 : How To Find Solutions To Differential Equations

Find the derivative with respect to x for the following function:

$y=e^{sin(3x)}+(x^5-3x)^3$

Possible Answers:

$\frac{dy}{dx}=3sin(3x)e^{sin(3x)-1}+(12x^4-9)(x^5-3x)^2$

$\frac{dy}{dx}=-3cos(3x)e^{sin(3x)}+(12x^4-9)(x^5-3x)^2$

$\frac{dy}{dx}=3cos(3x)e^{sin(3x)}+(15x^4-9)(x^5-3x)^2$

$\frac{dy}{dx}=-3sin(3x)e^{sin(3x)-1}+(12x^4-9)(x^5-3x^2)^2$

$\frac{dy}{dx}=3cos(3x)e^{sin(x)}+3(x^5-3x)^2$

Correct answer:

$\frac{dy}{dx}=3cos(3x)e^{sin(3x)}+(15x^4-9)(x^5-3x)^2$

Explanation:

Computation of the derivative requires use of the Chain Rule.

$y=e^{sin(3x)}+(x^5-3x)^3$

$\frac{dy}{dx}=\frac{d}{dx}[e^{sin(3x)}+(x^5-3x)^3)]$

For:

$y=e^{f(u)}$

$\frac{dy}{dx}=e^{f(u)}*f'(u)$

Here f(u)=sin(3x) , therefore, f'(u)=3cos(3x)

For:

y=(f(x))^a

$\frac{dy}{dx}=a*(f(x))^{a-1}*f'(x)$

Applying these two rules results in:

$\frac{dy}{dx}=e^{sin(3x)}(3cos(3x))+3(x^5-3x)^2(5x^4-3)$

Simplifying this equation results in one of the answer choices:

$\frac{dy}{dx}=3cos(3x)e^{sin(3x)}+(15x^4-9)(x^5-3x)^2$

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Example Question #70 : How To Find Solutions To Differential Equations

Solve the differential equation:

$\frac{dy}{dx}=\frac{2sinx+cosx}{y}$

Possible Answers:

$y=\sqrt{4cosx-2sinx+c}$

$y=\sqrt{8cosx-4sinx+c}$

$y=\sqrt{2sinx-4cosx+c}$

y=4cosx-2sinx+c

Correct answer:

$y=\sqrt{2sinx-4cosx+c}$

Explanation:

This is a separable equation, meaning to solve we want to separate, then integrate.

Step 1: Separate by putting the y components on one side of the equation, and x components on the other.

ydy=(2sinx+cosx)dx

Step 2: Integrate both sides of the equation.

$\int ydy= \int (2sinx+cosx)dx$

This gives us the following:

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

Step 3: Now to simplify we can solve for y.

$y^{2}=2sinx-4cosx+c$

$y=\sqrt{2sinx-4cosx+c}$

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Calculus 1 : Differential Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #61 : How To Find Solutions To Differential Equations

Example Question #62 : How To Find Solutions To Differential Equations

Example Question #63 : How To Find Solutions To Differential Equations

Example Question #64 : How To Find Solutions To Differential Equations

Example Question #65 : How To Find Solutions To Differential Equations

Example Question #66 : How To Find Solutions To Differential Equations

Example Question #67 : How To Find Solutions To Differential Equations

Example Question #68 : How To Find Solutions To Differential Equations

Example Question #69 : How To Find Solutions To Differential Equations

Example Question #70 : How To Find Solutions To Differential Equations

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