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Calculus 1 : Differential Equations

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Example Question #41 : How To Find Solutions To Differential Equations

Find the general solution for the following differential equation:

$3x^2-3\frac{dy}{dx}=2$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

First we must rearrange this separable differential equation so that we can place alone on one side and on the other side with the terms involving and any constants. We then integrate each side with respect to the appropriate variable and solve the result for to find the general solution of the differential equation:

$3x^2-3\frac{dy}{dx}=2$

$-3\frac{dy}{dx}=-3x^2+2$

$\frac{dy}{dx}=x^2-\frac{2}{3}$

$dy=(x^2-\frac{2}{3})dx$

$\int dy=\int \left(x^2-\frac{2}{3}\right)dx$

Remember when integrating, we increase the exponent by one and then divide the whole term by the value of the new exponent. Will we need to integrate each term that contains in this fashion.

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

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Example Question #42 : How To Find Solutions To Differential Equations

Find the general solution for the following differential equation:

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

Possible Answers:

$y=\pm\sqrt{\frac{1}{2}x^3+\frac{3}{2}x+C}$

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

$y=\pm\sqrt{\frac{2}{3}x^3-2x+C}$

$y=\pm\sqrt{\frac{1}{3}x^3+x+C}$

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

Correct answer:

$y=\pm\sqrt{\frac{2}{3}x^3-2x+C}$

Explanation:

First we must multiply to the right side of the equation so we have the terms with and the terms with . We can then integrate each side with respect to the appropriate variable and then solve the result for to find the general solution for the differential equation:

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

(y)dy=(x^2-1)dx

$\int (y)dy=\int (x^2-1)dx$

Remember when integrating, we increase the exponent by one and then divide the term by that new exponent value. We will do this with each term on both sides.

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

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

$y=\pm \sqrt{\frac{2}{3}x^3-2x+C}$

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Example Question #43 : How To Find Solutions To Differential Equations

Determine the general solution to the following differential equation:

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

Possible Answers:

$y=e^{x^2}+C$

$y=Ce^{2x}$

$y=e^{2x}+C$

y=2x+C

$y=Ce^{x^2}$

Correct answer:

$y=Ce^{x^2}$

Explanation:

In order to separate the differential equation such that is with the terms and is with the terms, we must divide both sides of the equation by and then multiply both sides by . We can then integrate each side with respect to the appropriate variable, and then solve for to find the general solution to the differential equation:

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

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

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

$\ln (y)=x^2+C$

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

$y=e^{x^2+C}$

$y=e^Ce^{x^2}$

$y=Ce^{x^2}$

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Example Question #44 : How To Find Solutions To Differential Equations

Determine the general solution to the following differential equation:

$\frac{dy}{dx}=\frac{x^3}{y}$

Possible Answers:

$y=\pm \sqrt{\frac{1}{2}x^4+C}$

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

$y=\pm \sqrt{\frac{2}{3}x^3+C}$

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

$y=\pm \sqrt{\frac{1}{3}x^3+C}$

Correct answer:

$y=\pm \sqrt{\frac{1}{2}x^4+C}$

Explanation:

For this separable differential equation, we can see that if we cross multiply we will have the term with , and the term with . After getting the equation into this form, we can integrate each side with respect to the appropriate variable, and then solve for to find the general solution to the differential equation:

$\frac{dy}{dx}=\frac{x^3}{y}$

(y)dy=(x^3)dx

$\int (y)dy=\int (x^3)dx$

Remember when integrating, we increase the exponent by one and then divide the term by that new exponent value. We will do this with each term on both sides.

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

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

$y=\pm \sqrt{\frac{1}{2}x^4+C}$

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Example Question #45 : How To Find Solutions To Differential Equations

Find the particular solution for the following initial value problem:

$\frac{dy}{dx}+2x=3x^2$

y(0)=2

Possible Answers:

y=3x^3-2x^2+2

y=x^3-x^2

y=3x^2-2x

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

y=x^3-x^2+2

Correct answer:

y=x^3-x^2+2

Explanation:

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:

$\frac{dy}{dx}+2x=3x^2$

$\frac{dy}{dx}=3x^2-2x$

dy=(3x^2-2x)dx

$\int dy=\int (3x^2-2x)dx$

Remember when integrating, we can use the power rule. This means to increase the exponent by one and then divide the term by that new exponent value. We will do this with each term on both sides.

y=x^3-x^2+C

Once we have found the general solution we plug in the initial conditions to solve for C.

$y(0)=2\rightarrow (0)^3-(0)^2+C=2\rightarrow C=2$

y=x^3-x^2+2

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Example Question #46 : How To Find Solutions To Differential Equations

Find the particular solution for the following differential equation:

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

y(0)=0

Possible Answers:

$y=\pm\sqrt{ 2x^2+2x+2}$

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

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

$y=\pm\sqrt{ \frac{1}{2}x^2+2x+\frac{1}{2}}$

$y=\pm\sqrt{ 2x^2+2x}$

Correct answer:

$y=\pm\sqrt{ 2x^2+2x}$

Explanation:

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. For this problem we can see that the desired arrangement is achieved by simply cross multiplying the differential equation. 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:

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

(y)dy=(2x+1)dx

$\int (y)dy=\int (2x+1)dx$

Remember when integrating, we can use the power rule. The power rule states to increase the exponent by one and then divide the term by that new exponent value. We will do this with each term on both sides.

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

y^2=2x^2+2x+C

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

Now that we have the general solution we can plug in the initial values and solve for C.

$y(0)=0\rightarrow \pm \sqrt{2(0)^2+2(0)+C}=0\rightarrow C=0$

$y=\pm\sqrt{ 2x^2+2x}$

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Example Question #47 : How To Find Solutions To Differential Equations

Find the first derivative of the following:

x^2e^x

Possible Answers:

2xe^x

$2e^{x^2}$

e^x(x^2+4x+2)

e^xx(x+2)

e^x(x+2)

Correct answer:

e^xx(x+2)

Explanation:

This problem requires the use of the product rule.

The product rule tells us that the derivative of,

$f(x)\cdot g(x)$ is f'(x)g(x)+f(x)g'(x) .

In this problem, f(x) is x^2 , and g(x) is e^x .

Therefore,

f'(x) = 2x , and g'(x) = e^x , so

f'(x)g(x)+f(x)g'(x) = 2xe^x+x^2e^x , and factoring out xe^x gives us xe^x(x+2) .

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Example Question #48 : How To Find Solutions To Differential Equations

Determine the general solution of y''+y'-6=0 , if possible.

Possible Answers:

y=(C_1+C_2)(e^-^3^x+e^2^x)

$\textup{There is no solution.}$

y=e^-^3^x+e^2^x+C

y=C_1e^-^3^x+C_2e^2^x

y=-3e^C^__1^x+2e^C^__2^x

Correct answer:

y=C_1e^-^3^x+C_2e^2^x

Explanation:

To find the general solution of y''+y'-6=0 , rewrite this in the form of its characteristic equation. This will be a quadratic.

r^2+r-6=0

Factorize and solve for the roots.

r^2+r-6=(r+3)(r-2)

r_1_,_2=-3,2

Since the roots are distinct, write the specific form of the general solution stated by one of the three homogenous equation rules.

y=C_1e^r^__1^x+C_2e^r^__2^x

Substitute the roots in the formula to obtain the general solution.

y=C_1e^-^3^x+C_2e^2^x

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Example Question #49 : How To Find Solutions To Differential Equations

Determine the general solution of y''-9=0 , if possible.

Possible Answers:

y=(C_1+C_2)(e^3^x+e^-^3^x)

y=(C_1+C_2)(e^3^x)

y=C_1e^-^3^x+C_2e^3^x

$\textup{The answer is not given.}$

y=-3e^C^__1^x+-3e^C^__2^x

Correct answer:

y=C_1e^-^3^x+C_2e^3^x

Explanation:

To solve for the general solution of y''-9=0 , convert this to its characteristic equation and solve for the roots.

r^2-9=0

r^2-9=(r-3)(r+3)=0

$r_1_,_2=\pm3$

Write the specific form of the general solution stated by one of the three homogenous equation rules for distinct roots.

y=C_1e^r^__1^x+C_2e^r^__2^x

Substitute the roots to obtain the general solution.

y=C_1e^-^3^x+C_2e^3^x

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Example Question #50 : How To Find Solutions To Differential Equations

Let y=x^x . Find $\frac{dy}{dx}$ .

(Hint: Use differential equations.)

Possible Answers:

$y'=x^{x-1}$

$y'=\frac{1}{1+\ln{x}}$

$y'=\frac{x^x}{\ln x}$

$y'=\ln{x} \cdot x^x$

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

Correct answer:

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

Explanation:

Let y=x^x . Find $\frac{dy}{dx}$ .

(Hint: Use differential equations.)

Start with y=x^x

First, take the natural logarithm of both sides:

$\ln{y}=\ln{(x^x)}$

Simplify:

$\ln{y}=x\ln{(x)}$

Take the derivatives:

$\frac{y'}{y}=\ln{(x)}+1$

Multiply both sides by :

$y'=(\ln{(x)}+1)y$

Substitute the original equation for :

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

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Calculus 1 : Differential Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #41 : How To Find Solutions To Differential Equations

Example Question #42 : How To Find Solutions To Differential Equations

Example Question #43 : How To Find Solutions To Differential Equations

Example Question #44 : How To Find Solutions To Differential Equations

Example Question #45 : How To Find Solutions To Differential Equations

Example Question #46 : How To Find Solutions To Differential Equations

Example Question #47 : How To Find Solutions To Differential Equations

Example Question #48 : How To Find Solutions To Differential Equations

Example Question #49 : How To Find Solutions To Differential Equations

Example Question #50 : How To Find Solutions To Differential Equations

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