# Verify Solutions of DEs Exercises Worked

#### Contents

- 1. {$ y=ce^{2x}\;; y'=2y $}
- 2. {$ y={x^2\over3}+{c\over x}\;; xy'+y=x^2 $}
- 3. {$ y={1\over2}+ce^{-x^2}\;; y'+2xy=x $}

#### Exercises

#### Worked 2

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

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

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

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

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

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