A differential equation (or "DE") contains derivatives or differentials. Show
Our task is to solve the differential equation. This will involve integration at some point, and we'll (mostly) end up with an expression along the lines of "y = ...". Recall from the Differential section in the Integration chapter, that a differential can be thought of as a derivative where `dy/dx` is actually not written in fraction form. Examples of Differentials
Examples of Differential EquationsExample 1We saw the following example in the Introduction to this chapter. It involves a derivative, `dy/dx`:
As we did before, we will integrate it. This will be a general solution (involving K, a constant of integration). So we proceed as follows:
and this gives
But where did that dy go from the `(dy)/(dx)`? Why did it seem to disappear? In this example, we appear to be integrating the x part only (on the right), but in fact we have integrated with respect to y as well (on the left). DEs are like that - you need to integrate with respect to two (sometimes more) different variables, one at a time. We could have written our question only using differentials:
(All I did was to multiply both sides of the original dy/dx in the question by dx.) Now we integrate both sides, the left side with respect to y (that's why we use "dy") and the right side with respect to x (that's why we use "dx") :
Then the answer is the same as before, but this time we have arrived at it considering the dy part more carefully:
On the left hand side, we have integrated `int dy = int 1 dy` to give us y. Note about the constant: We have integrated both sides, but there's a constant of integration on the right side only. What happened to the one on the left? The answer is quite straightforward. We do actually get a constant on both sides, but we can combine them into one constant (K) which we write on the right hand side. Example 2This example also involves differentials:
We have:
To solve this, we would integrate both sides, one at a time, as follows:
We have integrated with respect to θ on the left and with respect to t on the right. Here is the graph of our solution, taking `K=2`: Solving a differential equationFrom the above examples, we can see that solving a DE means finding an equation with no derivatives that satisfies the given DE. Solving a differential equation always involves one or more integration steps. It is important to be able to identify the type of DE we are dealing with before we attempt to solve it. DefinitionsFirst order DE: Contains only first derivatives Second order DE: Contains second derivatives (and possibly first derivatives also) Degree: The highest power of the highest derivative which occurs in the DE. Example 3 - Order and Degreea) `(d^2y)/(dx^2)+((dy)/(dx))^3-3x+2y=8` This DE has order 2 (the highest derivative appearing is the second derivative) and degree 1 (the power of the highest derivative is 1.) b) `((dy)/(dx))^5-2x=3 sin(x)-sin(y)` This DE has order 1 (the highest derivative appearing is the first derivative) and degree 5 (the power of the highest derivative is 5.) c) `(y'')^4+2(y')^7-5y=3` This DE has order 2 (the highest derivative appearing is the second derivative) and degree 4 (the power of the highest derivative is 4.) General and Particular SolutionsWhen we first performed integrations, we obtained a general solution (involving a constant, K). We obtained a particular solution by substituting known values for x and y. These known conditions are called boundary conditions (or initial conditions). It is the same concept when solving differential equations - find general solution first, then substitute given numbers to find particular solutions. Let's see some examples of first order, first degree DEs. Example 4a. Find the general solution for the differential equation `dy + 7x dx = 0` b. Find the particular solution given that `y(0)=3`. Answer (a) We simply need to subtract 7x dx from both sides, then insert integral signs and integrate:
NOTE 1: We are now writing our (simple) example as a differential equation. Earlier, we would have written this example as a basic integral, like this: `(dy)/(dx)+7x=0` Then `(dy)/(dx)=-7x` and so `y=-int7x dx=-7/2x^2+K` The answer is the same - the way of writing it, and thinking about it, is subtly different.
We could also have:
and so on. We'll come across such integrals a lot in this section. (b) We now use the information y(0) = 3 to find K. The information means that at x = 0, y = 3. We substitute these values into the equation that we found in part (a), to find the particular solution. `3=7/2(0)^2+K` gives K = 3. So the particular solution is: `y=-7/2x^2+3`, an "n"-shaped parabola. Here is the graph of the particular solution we just found: Example 5Find the particular solution of
given that when `x=0, y=2`. Answer We can write
as a differential equation:
Integrating both sides gives:
Applying the boundary conditions: x = 0, y = 2, we have K = 2 so:
Example 6Find the particular solution of
given that:
Answer Since y''' = 0, when we integrate once we get:
Integrating again gives:
Once more:
The boundary conditions are:
We need to substitute these values into our expressions for y'' and y' and our general solution, `y = (Ax^2)/2 + Bx + C`. Now
and
Finally,
So the particular solution for this question is:
Checking the solution by differentiating and substituting initial conditions:
Our solution is correct. Example 7After solving the differential equation,
(we will see how to solve this DE in the next section Separation of Variables), we obtain the result
Did we get the correct general solution? Answer Now, if `y=c ln x`, then `(dy)/(dx)=c/x` [See Derivative of the Logarithmic Function if you are rusty on this.) So `"LHS"=(dy)/(dx)ln x-y/x`
We conclude that we have the correct solution. Second Order DEsWe include two more examples here to give you an idea of second order DEs. We will see later in this chapter how to solve such Second Order Linear DEs. Example 8The general solution of the second order DE
is
Example 9The general solution of the second order DE
is
If we have the following boundary conditions:
then the particular solution is given by:
Example 10 - Second Order DEShow that
is a general solution for the differential equation
Answer We have a second order differential equation and we have been given the general solution. Our job is to show that the solution is correct. We do this by substituting the answer into the original 2nd order differential equation. We need to find the second derivative of y:
First derivative:
Second derivative:
Now for the check step: `"LHS"=(d^2y)/(dx^2)+4y`
Example 11 - Second Order DEShow that `(d^2y)/(dx^2)=2(dy)/(dx)` has a solution of y = c1 + c2e2x Answer Since
and
It is obvious that .`(d^2y)/(dx^2)=2(dy)/(dx)` Problem SolverNeed help solving a different Calculus problem? Try the Problem Solver. Disclaimer: IntMath.com does not guarantee the accuracy of results. Problem Solver provided by Mathway. What is the general solution of a differential equation?The general solution to a differential equation is the most general form that the solution can take and doesn't take any initial conditions into account. Example 5 y(t)=34+ct2 y ( t ) = 3 4 + c t 2 is the general solution to 2ty′+4y=3.
What is general solution of differential equation with example?The general solution geometrically represents an n-parameter family of curves. For example, the general solution of the differential equation \frac{dy}{dx} = 3x^2, which turns out to be y = x^3 + c where c is an arbitrary constant, denotes a one-parameter family of curves as shown in the figure below.
How do you find the general solution of a system of equations?Given a system of equations, the steps for writing out the general solution are:. Row reduce the augmented matrix for the system.. Write out the equations from the row-reduced matrix.. Solve for the variables that have a leading one in their column.. Label the remaining variables as free variables.. How do you find the general solution of a partial differential equation?Since the constants may depend on the other variable y, the general solution of the PDE will be u(x, y) = f(y) cosx + g(y) sinx, where f and g are arbitrary functions. To check that this is indeed a solution, simply substitute the expression back into the equation. ux = f(x).
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