Differential Equations?

Question

Solve the differential equation y'=y-x+1 to find the value of y, when x=1 (full working apprieciated)

Answer 1

You won't be able to find a specific value of y when x=1 unless you have some other initial condition that your y has to satisfy.

Other than this consideration, this is a *simple* homework problem.

Answer 2

Michaelsgdec, the correct solution is y=x+Ce^x, not y=x+C.

telemarchus, what you have here is, as mentioned, a first-order linear differential equation. This is good, because there is a method of solving these equations that always works, which makes them much more convenient than a lot of other differential equations. The first thing you should do is shuffle all your y terms over to one side of the equation, like so:

y'-y=1-x

You may recognize this as the canonical form of linear first-order ODEs, which should be given in your textbook as y'+f(x)y=g(x) or something similar. In this case, f(x)=-1, but it is free to be any function of x and this method will still work. What we need to do is to find a function that, when multiplied by f(x), will yield its own derivative. Such a function is called an integrating factor. Normally when solving a problem like this, I would simply note that e^(-x) fulfills the requirements and move on, however in this case I shall show you how to get that solution so that you will remember how to do it when f(x) is more complicated. What we do is we set up the requirements of the integrating factor as a second differential equation, like so:

h(x)f(x)=h'(x)

We then divide by h(x):

h'(x)/h(x)=f(x)

Now hopefully you remember from your calculus classes the integral of dx/x is simply ln x. So we can integrate both sides of this equation to get:

ln h(x) = ∫f(x)

Exponentiating, we have:

h(x)=e^(∫f(x))

Which is our integrating factor. Let's try this on the equation we're working with right now, which is:

y'-y=1-x

Thus we want:

h'(x)=-h(x)

Divide by h(x):

h'(x)/h(x)=-1

Integrate:

ln h(x) = -x

Exponentiate:

h(x) = e^(-x)

Which was the integrating factor I suggested. Now if you were paying really close attention you may have noticed something missing: specifically, the arbitrary constant of integration. Normally, when finding an antiderivative, we add an arbitrary constant to make note of the fact that every function has infinitely many antiderivatives, all of which are valid solutions to the problem. The reason I didn't do that here is because I am not interested in finding all of the integrating factors I could use, I only want to find one, and picking the one where C=0 is as good as any other. It can be shown (though I will not do so here) that if I were to use a different integrating factor, such as e^(-x+22) or e^(-x-13), or in general e^(-x+C), we would eventually get the same solution to the problem, because C would be absorbed by another arbitrary constant of integration later in the problem. I mention this, because a lot of students get into the habit of thinking that adding + C at the end is just an annoying notational convention, and I thought it would help to illustrate when that constant is actually important by giving a situation when it isn't (namely, when we explicitly care only about getting one solution to a problem and not the most general solution possible).

But going back to the differential equation itself:

y'-y=1-x

Now that we've found a suitable integrating factor, we multiply the equation by that factor, thus we have:

e^(-x)y' - e^(-x)y = (1-x)e^(-x)

Now why did we do this? Let's look at the general form for a moment:

y' + f(x)y = g(x)

Multiplied by the integrating factor we have:

h(x)y' + h(x)f(x)y = h(x)g(x)

But by the definition of our integrating factor, h(x)f(x)=h'(x). Thus:

h(x)y' + h'(x)y = h(x)g(x)

Hm... hy' + h'y... doesn't that look a lot like the product rule? Indeed, it IS the product rule, which means that we can rewrite the equation as:

(h(x)y)' = h(x)g(x)

And now note that instead of having seperate y and y' terms, we only have one y term, which we can isolate through the ordinary techniques of calculus and algebra. This is what the integrating factor does: it allows us to combine the y and y' terms so we can safely integrate them. Applying this logic to our original differential equation we have:

(e^(-x)y)' = (1-x)e^(-x)

Now we integrate both sides. In this case, since we want the most general solution to the differential equation itself, we WILL have to keep track of constants of integration. So we have:

∫(e^(-x)y)' = ∫(1-x)e^(-x)
e^(-x)y + C = ∫e^(-x) - ∫xe^(-x)
Applying integration by parts to the right integral:
e^(-x)y + C = ∫e^(-x) - (-xe^(-x) - ∫-e^(-x))
e^(-x)y + C = ∫e^(-x) + xe^(-x) - ∫e^(-x)
e^(-x)y + C = xe^(-x) + K
And since one arbitrary constant is as good as another, we can simply call (K-C) C, thereby absorbing the two constants into one. This gives us:

e^(-x)y = xe^(-x) + C

Dividing by e^(-x) gives us our final solution:

y=x + Ce^x

Check of answer: y'=1 + Ce^x. Substitute. Does 1 + Ce^x = x + Ce^x - x + 1? Indeed, it does. Therefore our solution is correct.

Now, the problem asks you to find the value of y when x=1. If we had even a single value for y we could figure out what C is and thereby be able to determine the value of y at x=1. But since we don't, all we can do is give the general form of the equation. Note the importance of that C here - suppose that a student carelessly omitted it and was asked to solve an initial value problem with the same equation you had and say... y(0)=2. They would be unable to find an equation that fits, although clearly there is one, given by C=2. Thus, when solving differential equations, ensuring that you find ALL the antiderivatives of a given function is actually kind of important.

Answer 3

it is linear
dy/dx-y=x+1

integral factor is e raise to the integral of 1 is equal to e raise to the (x).

solution
y times the integral factor = integral of x+1 times e^x+ c

y(e^x)= xe^x -e^x+e^x+c

ye^x=xe^x+c the general solution

u can't find this cause the c or arbituary constant is unknown,
U shold pick the value of point x and y so we will determine the particular solution.

Answer 4

y = INT[f*e^G] / e^G
g= -1
G= -x
f= -x+1
y = INT[(-x+1)*e^-x] / (e^-x)
y = (-INT[xe^-x]+INT[e^-x]) / (e^-x)
y = (e^-x)(x+1-1) / (e^-x)
y = x + C

for explaination see http://answers.yahoo.com/question/index;_ylt=AmBUtUCFiPXBmL_qMZHoPDrsy6IX?qid=20060811102603AAQC4su

man, you copied the question wrong, the RHS is -x+1, not x+1

Answer 5

the answer is unknown

Answer 6

some mistakes r there