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# Verify Solutions of DEs Exercises Worked

Worked Exercises

## Exercises

Verify that the function is a solution of the differential equation on some interval, for any choice of the arbitrary constants appearing in the function

1. {$$\large y=ce^{2x}\;; \quad y'=2y$$}

{\large \begin{align} {d\over{dx}} \left(ce^{2x}\right) \quad & ? \quad 2 \left(ce^{2x} \right) \cr c{d\over{dx}}(2x){d\over{dx}}\left(e^{2x}\right) \quad & ? \quad 2ce^{2x} \cr c2e^{2x} \quad & ? \quad 2ce^{2x} \cr \therefore 2ce^{2x} & = 2ce^{2x} \end{align}}

Some solutions of {$y^\prime = 2y$}. Isoclines are shown in light magenta, and the slope or direction field is shown as short blue lines. Some solutions of the DE are show with thick colored lines.

2. {$$\large y={x^2\over3}+{c\over x}\;; \quad xy'+y=x^2$$}

Some solutions of {$xy'+y=x^2$}. Isoclines were originally entered in tan, but are almost completely obscured by the slope field shown with short light blue lines. Notice that some parts of solutions cross the same isocline twice near x = 0. This seems to indicate that those parts have a point of inflection on x = 0.

To demonstrate the proposed solution, it is substituted in the differential equations.

{\large \begin{align} x\left({x^2\over{3}} + {c\over{x}} \right)^\prime + \left({x^2\over{3}} + {c\over{x}} \right) \quad & ? \quad x^2 \cr x\left({2\over{3}}x - {1\over{x^2}}c \right) + {1\over{3}}x^2 + {1\over{x}}c \quad & ? \quad x^2 \cr {2\over{3}}x^2 - {1\over{x}}c + {1\over{3}}x^2 + {1\over{x}}c \quad & ? \quad x^2 \cr \therefore x^2 & = x^2 \end{align}}

Finding the isoclines:

Isoclines are lines through points where the slope of a function are the same. The first derivative of a function is a function for slope of the given function. So when the first derivative is set equal to a number, the result is an equation for an isocline.

{\large \begin{align} xy'+y & = x^2 \cr y & = x^2 - xy' \end{align}}

Generally, we use the parameter m (or c) for the slope, which is just y'. Then for every m we chose, we have the equation for an isocline.

{$$\large y = x^2 - xm$$}

Isoclines give an overall impression of the slope field. But they are much easier to understand if we sketch in little segments with the slope m on the isocline. This is probably the best way to proceed if one is sketching the slope field by hand.

Of course we could just find m for various points on the plain by solving the equation for m, and plugging in the x and y coordinates for the various points.

{$$\large x - {y\over{x}} = m$$}

This is the essential meaning of a field. There is a value of m for every point on the plane (except in this case for points that lie on the y axis). It is a slope field because the value represent slopes. It would be very messy to try to write a number on the plane for every selected point (and impossible to write a number for every point where m is defined), so we just put little lines with the appropriate slopes at selected points. This provides a very intuitive picture of the slope field.

3. {$$\large y={1\over2}+ce^{-x^2}\;; \quad y'+2xy=x$$}

Slope field for {$y^\prime + 2xy = x$}.

{\large \begin{align} \left({1\over2}+ce^{-x^2}\right)^\prime + 2x\left({1\over2}+ce^{-x^2}\right) \quad & ? \quad x \cr \left({1\over2}\right)^\prime +\left(ce^{-x^2}\right)^\prime + x + 2xce^{-x^2} \quad & ? \quad x \cr c(-2x)e^{-x^2} + x + 2xce^{-x^2} \quad & ? \quad x \cr \therefore x & = x \end{align}}

Sources:

Trench, William F., "Elementary Differential Equations with Boundary Value Problems" (2013). Faculty Authored Books. 9. Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.

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