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Calculus 1 : Equations

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Example Questions

Example Question #271 : Equations

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

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

Possible Answers:

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

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

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

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

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 #22 : 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:

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

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

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

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

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 #23 : 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:

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

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

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

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

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 #272 : Equations

Differentiate the polynomial.

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

Possible Answers:

$2x^{2}+4x$

4x+1

2x+4

x+1

$x^{2}+1$

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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Example Question #31 : Differential Equations

Differentiate the expression.

$e^{6x}$

Possible Answers:

6xe

$6e^{6x}$

$e^{6x}$

Correct answer:

$6e^{6x}$

Explanation:

We will use the fact that $e^{au} = ae^{au}$ to differentiate. Let a=6 and u=x . Substituing our values we can see the derivative will be $6e^{6x}$ .

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Example Question #272 : Equations

Differentiate the expression.

$x^{2}\sin(x)$

Possible Answers:

$2x\sin(x)$

x^2

$x (2 \sin(x)+x \cos(x))$

$2x\cos(x)$

$x\cos(x)$

Correct answer:

$x (2 \sin(x)+x \cos(x))$

Explanation:

Using the product rule, we determine the derivative of $f(x)\cdot g(x) = {f}'(x)\cdot g(x) + f(x)\cdot {g}'(x))$
Let $f(x) = x^{2}$ and $g(x) = \sin(x)$ . We can see that f{}'(x) = 2x and $g{}'(x) = \cos(x)$ .

Plugging in our values into the product rule formula, we are left with the final derivative of $x (2 \sin(x)+x \cos(x))$ .

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Example Question #271 : Equations

Differentiate the value.

Possible Answers:

74x

$74x^{2}$

Correct answer:

Explanation:

According to the power rule, whenever we differentiate a constant value it will reduce to zero. Since the only term of our function is a constant, we can only differentiate $74 \rightarrow 0$ .

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Example Question #272 : Equations

Find {f}'(x) .

$f(x) = e^{2x^{2}}$

Possible Answers:

$4xe^{2x^{2}}$

$e^{4x}$

$2e^{x^{2}}$

$4e^{x}$

$e^{2x^{2}}$

Correct answer:

$4xe^{2x^{2}}$

Explanation:

Using the chain rule, we will differentiate the exponent of our exponential function, and then multiply our original function. Differentiating our exponent with the power rule will yield . Using the chain rule we will multiply this by our original function resulting in $4xe^{2x^{2}}$ .

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Example Question #273 : Equations

Find {f}'(x) .

$f(x) = x^{2}-4$

Possible Answers:

2x-4

$x^{2}$

(x+2)(x-2)

Correct answer:

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 is a constant value, so according to the power rule this term will become .

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Example Question #31 : Differential Equations

Differentiate the logarithm.

$\ln(4x)$

Possible Answers:

$\frac{1}{x}$

$\frac{1}{4x}$

$4\ln(4x)$

Correct answer:

$\frac{1}{x}$

Explanation:

Using the chain rule, we will determine the derivative of our function will be $\frac{d\log(u)}{du}\cdot \frac{du}{dx}$ .

The derivative of the log function is $\frac{1}{x}$ , and our second term of the chain rule will cancel out $\frac{1}{4x}\cdot 4=\frac{4}{4x}=\frac{1}{x}$ .

Thus our derivative will be $\frac{1}{x}$ .

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Calculus 1 : Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #271 : Equations

Example Question #22 : Differential Equations

Example Question #23 : Differential Equations

Example Question #272 : Equations

Example Question #31 : Differential Equations

Example Question #272 : Equations

Example Question #271 : Equations

Example Question #272 : Equations

Example Question #273 : Equations

Example Question #31 : Differential Equations

All Calculus 1 Resources

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