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Calculus 1 : Solutions to Differential Equations

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Example Question #11 : Solutions To Differential Equations

Find f'(2) of the following equation:

f(x)=x^3+4x^2-7x-9

Possible Answers:

Correct answer:

Explanation:

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

f(x)=x^3+4x^2-7x-9

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

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

Therefore the derivative of our function is:

f'(x)=3x^2+8x-7

f'(2)=3(2)^2+8(2)-7=21

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Example Question #12 : Solutions To Differential Equations

Find f'(1) for the following equation:

f(x)=3x^5(2x+4)

Possible Answers:

Undefined

Correct answer:

Explanation:

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,

h(x)=3x^5 and g(x)=2x+4

h'(x)=15x^4 and g'(x)=2

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

f(x)=3x^5(2x+4)

f'(x)=15x^2(2x+4)+3x^5(2)

Now plug in x=1 and solve.

f'(1)=15(1)^2(2(1)+4)+3(1)^5(2)=96

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Example Question #13 : Solutions To Differential Equations

Find the solution to the following equation at f'(x)=-2

f(x)=4x^3-6x+1

Possible Answers:

Undefined

Correct answer:

Explanation:

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,

f(x)=4x^3-6x+1

f'(x)=12x^2-6

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

f'(-2)=12(-2)^2-6=48-6=42

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Example Question #14 : Solutions To Differential Equations

Find f'(3) for the following equation:

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

Possible Answers:

$\ln(3)+6$

$23\frac{1}{3}$

$\ln(3)+23$

$\frac{23}{3}$

Correct answer:

$23\frac{1}{3}$

Explanation:

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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Example Question #15 : Solutions To Differential Equations

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

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

Possible Answers:

$\frac{1}{y}=x^2+\frac{5}{4}$

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

$-\frac{3}{y^3}=x^2+\frac{11}{4}$

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

Correct answer:

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

Explanation:

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:

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

The 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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Example Question #2341 : Calculus

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

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

Possible Answers:

$-\frac{1}{y}=2x^3+\frac{1}{8}$

$-\frac{1}{3y^3}=2x^3+\frac{259}{12}$

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

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

Correct answer:

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

Explanation:

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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Example Question #17 : Solutions To Differential Equations

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

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

Possible Answers:

$\sqrt{y}=\frac{2}{3}(x)^\frac{3}{2} -\frac{32}{3}$

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

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

$y=x^\frac{3}{2}$

Correct answer:

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

Explanation:

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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Example Question #18 : Solutions To Differential Equations

Find the particular solution given y(1)=0 .

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The first thing we must do is rewrite the equation:

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

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

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

c=-1

The particular solution is then:

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

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Example Question #11 : How To Find Solutions To Differential Equations

Find the particular solution given y(2)=1 .

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2x^3+3x^2-x+7}{y^5+3y^2-2y}$

Possible Answers:

$\frac{1}{6}y^6+y^3-y^2=\frac{1}{2}x^4+x^3-\frac{1}{2}x^2+7x-\frac{167}{6}$

$\frac{1}{6}y^6+y^3-y^2=\frac{1}{2}x^4+x^3-\frac{1}{2}x^2+7x$

$y^6+y^3-y^2=\frac{1}{8}x^4+x^3-\frac{1}{2}x^2-7$

$\frac{1}{6}y^6+y^3-y^2=\frac{1}{2}x^4+x^3-\frac{1}{2}x^2+7x-28$

Correct answer:

$\frac{1}{6}y^6+y^3-y^2=\frac{1}{2}x^4+x^3-\frac{1}{2}x^2+7x-\frac{167}{6}$

Explanation:

The first thing we must do is rewrite the equation:

$({y^5+3y^2-2y})dy=({2x^3+3x^2-x+7})dx$

We can then find the integrals:

$({y^5+3y^2-2y})dy=({2x^3+3x^2-x+7})dx$

The integrals are as follows:

$\int ({y^5+3y^2-2y})dy=\frac{1}{6}y^6+y^3-y^2$

$\int ({2x^3+3x^2-x+7})dx=\frac{1}{2}x^4+x^3-\frac{1}{2}x^2+7x$

We're left with

$\frac{1}{6}y^6+y^3-y^2=\frac{1}{2}x^4+x^3-\frac{1}{2}x^2+7x+c$

Plugging in the initial conditions and solving for c gives us:

$\frac{1}{6}+1-1=8-8-2+14+c$

$c=-\frac{167}{6}$

The particular solution is then,

$\frac{1}{6}y^6+y^3-y^2=\frac{1}{2}x^4+x^3-\frac{1}{2}x^2+7x-\frac{167}{6}$

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Example Question #20 : Solutions To Differential Equations

Differentiate the polynomial.

$x^{2}+4x+1$

Possible Answers:

$2x^{2}+4x$

x+1

4x+1

$x^{2}+1$

2x+4

Correct answer:

2x+4

Explanation:

Using the power rule, we can differentiate our first term reducing the power by one and multiplying our term by the original power. $x^{2}$ , will thus become . The second term , will thus become . The last term is a constant value, so according to the power rule this term will become .

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Calculus 1 : Solutions to Differential Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #11 : Solutions To Differential Equations

Example Question #12 : Solutions To Differential Equations

Example Question #13 : Solutions To Differential Equations

Example Question #14 : Solutions To Differential Equations

Example Question #15 : Solutions To Differential Equations

Example Question #2341 : Calculus

Example Question #17 : Solutions To Differential Equations

Example Question #18 : Solutions To Differential Equations

Example Question #11 : How To Find Solutions To Differential Equations

Example Question #20 : Solutions To Differential Equations

All Calculus 1 Resources

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