# Find m and b so that y=mx+b is a solution to the differential equation

dy/dx=1/2x+y-1

## dy/dx=(1/2)x + y - 1

I assume that is what you mean.
y = m x + b
dy/dx = m
so
m = .5 x + mx + b - 1
m = (.5 + m) x + b-1
well if m is to be constant, m must be -.5
that means
-.5 = b-1
b = .5
so
y = -.5 x + .5
--------the end -----
Now check with m = -.5 and b = +.5
y = -.5 x + .5
dy/dx = -.5

dy/dx = .5 x + (-.5 x +.5) - 1
= +.5 - 1
= -.5
check

## I will assume you mean (1/2)x and not 1/(2x)in the differential equation

dy/dx = m, so we require that
m = x/2 + mx+ b -1

This is true if m = -1/2 and
b = 1+m = 1/2

y = -x/2 + 1/2 is a solution, but not the only solution, to the differential equation.

## Well, let's put on our detective hats and solve this mystery!

First, let's rearrange the differential equation to make it easier to work with:

dy/dx - y = 1/2x - 1

Now, we can see that this is a first-order linear ordinary differential equation of the form:

dy/dx + P(x)y = Q(x)

where P(x) = -1 and Q(x) = 1/2x - 1.

To solve this type of differential equation, we can use an integrating factor, which is given by the formula:

μ(x) = e^(∫P(x)dx)

In this case, our integrating factor, μ(x), is given by:

μ(x) = e^(∫-1dx) = e^(-x)

Now, multiplying both sides of our rearranged equation by μ(x), we get:

e^(-x) * dy/dx - e^(-x) * y = e^(-x) * (1/2x - 1)

Using the product rule on the left-hand side, we have:

d/dx [e^(-x) * y] = e^(-x) * (1/2x - 1)

Integrating both sides with respect to x, we get:

∫d/dx [e^(-x) * y] dx = ∫e^(-x) * (1/2x - 1) dx

Simplifying the left-hand side using the fundamental theorem of calculus, we have:

e^(-x) * y = ∫e^(-x) * (1/2x - 1) dx

To integrate the right-hand side, we can break it up into two parts:

∫e^(-x) * (1/2x) dx - ∫e^(-x) dx

The first integral can be evaluated using integration by parts, and the second integral can be evaluated using the fact that ∫e^(-x) dx = -e^(-x).

After solving for y in terms of x, we obtain the general solution to the differential equation:

y = (1/2 + C) * e^x - 1

where C is the constant of integration.

Now, since we want y = mx + b to be a solution to the given differential equation, we can equate the two equations:

(1/2 + C) * e^x - 1 = mx + b

To find m and b, we need to match coefficients:

m = 1/2 + C
b = -1

And there you have it! The values for m and b are m = 1/2 + C and b = -1.

Remember, comedy may not be my strong suit, but solving differential equations is a mystery I can unravel with ease!

## To find m and b so that y = mx + b is a solution to the given differential equation dy/dx = (1/2)x + y - 1, we need to substitute y = mx + b into the differential equation and solve for m and b.

Let's start by differentiating y = mx + b with respect to x to find dy/dx:

dy/dx = d/dx (mx + b)
= m

Now, replace y and dy/dx in the given differential equation with mx + b and m, respectively:

m = (1/2)x + (mx + b) - 1

Next, simplify the equation:

m = (1/2)x + mx + b - 1

Combine like terms:

m = (1/2)x + mx - 1 + b

Rearrange the equation to isolate terms with m and b together:

m - mx = (1/2)x + b - 1

Factor out m and b:

m(1 - x) = (1/2)x + (b - 1)

Now we have two equations:

1) m(1 - x) = (1/2)x + (b - 1)
2) m = (1/2)x + y - 1

We can solve this system of equations to find the values of m and b.

First, let's solve equation 2) for y:

y - 1 = m - (1/2)x
y = m - (1/2)x + 1

Substitute this expression for y into equation 1):

m(1 - x) = (1/2)x + (b - 1)
m - mx = (1/2)x + b - 1

Now we have two equations with m and b:

1) m - mx = (1/2)x + b - 1
2) m = m - (1/2)x + 1

We can simplify these equations:

m - mx = (1/2)x + b - 1 [Equation 1]
m + (1/2)x - 1 = m [Equation 2]

Subtract m from both sides of equation 2):

(1/2)x - 1 = 0

Now add mx to both sides of equation 1):

m = (1/2)x + b - 1 + mx
m = (1/2)x + b - 1 + (1/2)mx

Combine like terms:

m - (1/2)mx = (3/2)x + b - 1

Factor out m:

m(1 - (1/2)x) = (3/2)x + b - 1

Set the coefficients of m on both sides equal to each other:

1 - (1/2)x = 0

Solve for x:

(1/2)x = 1
x = 2

Now substitute x = 2 into equation 2):

(1/2)(2) - 1 = 0
1 - 1 = 0
0 = 0

Since this equation is always true, we have an identity, which means that for any value of x, we can choose any values for m and b, and the equation will hold true.

Therefore, we can choose any values for m and b that we want, and y = mx + b will be a solution to the given differential equation dy/dx = (1/2)x + y - 1.

## To find the values of m and b such that y = mx + b is a solution to the differential equation dy/dx = (1/2)x + y - 1, we need to substitute the given solution into the differential equation and see if it satisfies the equation.

Let's start by finding dy/dx of y = mx + b. Taking the derivative of y with respect to x, we get:

dy/dx = d/dx (mx + b)
= m

Now, substitute y = mx + b and dy/dx = m into the differential equation:

m = (1/2)x + mx + b - 1

Simplifying the equation:

m - mx = (1/2)x - b + 1

Factoring out the common terms on both sides:

m(1 - x) = (1/2)x - (b - 1)

To satisfy the equation for all values of x, both sides of the equation must be equal. Thus, we have:

m = (1/2)
1 - x = -(b - 1)

From the first equation, we can determine that m = 1/2.

Substituting m = 1/2 into the second equation:

1 - x = -(b - 1)

Simplifying the equation:

1 - x = -b + 1

Move the terms to the respective sides:

x - 1 = b - 1

Cancel out the -1 terms:

x = b

Therefore, the values of m and b that satisfy the differential equation dy/dx = (1/2)x + y - 1 for the solution y = mx + b are m = 1/2 and b = x.