The x is 3 the y is 21 what is the constant of proportionality I need to know by tomorrow

Determine whether each first-order differential equation is separable, linear, both, or neither. 1. ????y????x+????xy=x2y2 2. y+

Mkey [24]

Answer:

a) Linear

b) Linear

c) Linear

d) Neither

See explanation below.

Step-by-step explanation:

a) $\frac{dy}{dx} +e^x y = x^2 y^2$

For this case the differential equation have the following general form:

$y' +p(x) y = q(x) y^n$

Where $p(x) =e^x$ and $q(x) = x^2$ and since n>1 we can see that is a linear differential equation.

b) $y + sin x = x^3 y'$

We can rewrite the following equation on this way:

$y' -\frac{1}{x^3} y= \frac{sin (x)}{x^3}$

For this case the differential equation have the following general form:

$y' +p(x) y = q(x) y^n$

Where $p(x) =-\frac{1}{x^3}$ and $q(x) = \frac{sin(x)}{x^3}$ and since n=0 we can see that is a linear differential equation.

c) $ln x -x^2 y =xy'$

For this case we can write the differential equation on this way:

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

For this case the differential equation have the following general form:

$y' +p(x) y = q(x) y^n$

Where $p(x) =x$ and $q(x) = \frac{ln(x)}{x}$ and since n=0 we can see that is a linear differential equation.

d) $\frac{dy}{dx} + cos y = tan x$

For this case we can't express the differential equation in terms:

$y' +p(x) y = q(x) y^n$

So the is not linear, and since we can separate the variables in order to integrate is not separable. So then the answer for this one is neither.

4 0

3 years ago

Provide conditions under which the mean function is linear but has a negative coefficient

den301095 [7]

Depending on your dependent/outcome variable, a negative value for your constant/intercept should not be a cause for concern. This simply means that the expected value on your dependent variable will be less than 0 when all independent/predictor variables are set to 0. For some dependent variables, this would be expected. For example, if the mean value of your dependent variable is negative, it would be no surprise whatsoever that the constant is negative; in fact, if you got a positive value for the constant in this situation, it might be cause for concern (depending on your independent variables).Even if your dependent variable is typically/always positive (i.e., has a positive mean value), it wouldn't necessarily be surprising to have a negative constant. For example, consider an independent variable that has a strongly positive relationship to a dependent variable. The values of the dependent variable are positive and have a range from 1-5, and the values of the independent variable are positive and have a range from 100-110. In this case, it would not be surprising if the regression line crossed the x-axis somewhere between x=0 and x=100 (i.e., from the first quadrant to the fourth quadrant), which would result in a negative value for the constant.The bottom line is that you need to have a good sense of your model and the variables within it, and a negative value on the constant should not generally be a cause for concern. Typically, it is the overall relationships between the variables that will be of the most importance in a linear regression model, not the value of the constant.

5 0

4 years ago

Can someone help me with this question ? 10points.

Anestetic [448]

It is the one on the farthest right

4 0

3 years ago

Select all sets to which the value belongs

Strike441 [17]

Answer:

$\sqrt{2} - \sqrt{2} = 0$