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For each value of the integration constant we get a 'double line' on which the product is zero. Categories The first two examples are linear equations - the third is non-linear. A non-linear equation contains some product of two expressions of the dependent variable y x. See whether you understand the difference: 1 d y x k x dx 2 d2 y x k x d x2 2. Methods There are two main groups of methods for solving differential equations: a the symbolic methods, leading to equations b the numerical methods, leading to numbers. In the examples above, we have used symbolic methods.

When these work, they are fine. They can give an overview that is difficult to obtain numerically.

However, most differential equations to not have a symbolic solution. If a solution exists, finding it often requires puzzling and patience. Numerical methods are more flexible. However, they also have their problems. To become a good differential-equation-solver you will need to master both methods.

We begin with the symbolic methods. Over the centuries many tricks have evolved to solve these. Separable Equation If each of the terms in a differential equation only contains a single variable, then the terms can be integrated directly. You can regard any particular solution as the intersection of the 3D function: 2 1.

This is one of the lines of constant height that you see in a contour plot. In this example these are circles around the origin. Integration Constant It is important to include the integration constant immediately on integration. This example shows why. Ratio Equation You can often bring a first order equation into a separable form using a substitution.

One group of equations where this works are the 'ratio' equations. These can be written in the form: y dy f dx x. The 'circles' example is a ratio equation, be it a very simple one. If you replace x and y in an equation by kx and ky, and the k's cancel in the result, then you have a ratio equation. These are brought into a separable form using the substitution: y z x.

It is not easy to simplify this, but the contour plot shows particular solutions. To get the plot, I had to reduce the x-range.

We can choose one of the two functions u x and v x. Here we choose u such that the first term in the last equation becomes zero: du u dx 0 du dx u ln u x uex. There is no need to include an integration constant here. The remaining equation becomes u dv x 2 dx dv x 2 e x dx x 2 x.

We will see other methods for linear equations when we consider second order equations. With partial differential quotients. Mathcad does not have symbols for these. If we start with this equation, how do we obtain the solution? We first check that it is indeed a differential. The two are indeed equal, so the equation is a differential. The cosine term occurs in both integrals because it contains both an x and a y.

The general solution is the combination of the two results not the sum! Do not use these. It is interesting to see how complicated the solutions of this simple-looking equation are.

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The Integrating Factor We take the differential equation from the previous paragraph, and divide it with some function here simply x : 1 y sin x y x sin x y dx dy 0 x x 1. These are not equal, so the new equation is no longer exact.

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## Differential Equations With Mathcad Prime

However, if we multiply equation 1 with x, we get back the original, exact, equation. Here, x is an integrating factor: a factor which transforms the equation into one that can be solved. It can be proven that all first order, first degree equations have integrating factors.

The problem is to find one of them. There is no general method for that.

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There are two groups of equations that can often be solved: 1 The equation is the product of two equations of a lower degree. As the parts belong to the same solution, there is only one integration constant. The solution is the line at zero height. As you see, the solution is a combination of a circle and a hyperbola.

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These are the solutions of the two parts of the equation. Below, I have made separate plots of the two solutions. The circles disappear for negative values of the integration constant; the hyperbola solutions switch from one branch to the other. This parametric plot shows that the solutions are parabolas that shift along the x-axis with increasing C.

There is one more solution: y 0 You can easily check this in the original equation. You cannot find it by giving C some value.

## Download Differential Models: An Introduction With Mathcad

This is a singular solution which is not the same as a particular solution. The x-axis is tangent to all particular solutions. We will look at a different example: dy 2 x dx. The solution is a cusp. Increasing C shifts the cusp upwards. As you can see, all points of the yaxis obey the differential equation. You cannot see this in the general solution, so this is again a singular solution.

The contour plot of this example is not interesting. This is a set of straight lines with varying origin and slope. The particular solutions are tangent lines of the envelope. The envelope is a singular solution. Note that there are no solutions inside the envelope. This equation is linear in x. We can eliminate p.

The others look like sheared parabolas.

Important exceptions are some of the linear equations. Linear equations have an important property: the sum of two solutions is also a solution. We then add a particular solution to include the effect of the right hand term. Even our simple equation cannot always be solved, but there are useful special cases where a solution can be obtained. Homogeneoous Equation The following function is a solution to the homogeneous equation: y x e k x d y x k dx kx.