These ten exercises follow the companion guide on differential equations. They run from classifying an equation by order and degree through direct integration, separable and linear first-order equations, the three cases of homogeneous second-order equations, a non-homogeneous equation, and an economic application. Work each one fully before reading the solutions, which explain the reasoning rather than just stating the answer.
Exercises
Exercise 1. State the order and the degree of the equation below, and explain why the fourth power does not make the degree four.
Exercise 2. Find the general solution of the equation below by direct integration.
Exercise 3. Using your general solution from Exercise 2, find the particular solution that satisfies the condition that y equals 3 when x equals 0.
Exercise 4. Solve the second-order equation below given that the first derivative equals 5 at x equals 0 and the function equals minus 1 at x equals 0.
Exercise 5. Solve the separable equation below given that y equals 3 when x equals 0.
Exercise 6. Solve the linear first-order equation below using an integrating factor.
Exercise 7. Find the general solution of the homogeneous equation below.
Exercise 8. Find the general solution of the homogeneous equation below, whose roots are complex.
Exercise 9. Find the general solution of the non-homogeneous equation below, using a complementary function and a particular integral.
Exercise 10. In a market, demand is 80 minus 3P and supply is minus 10 plus 2P, with an adjustment speed of 1, so the rate of price change equals the excess demand. Given that the price is 5 at time zero, find the price as a function of time and state the long-run equilibrium price.
Solutions
Solution 1. The order is the highest derivative present, which is the third derivative, so the order is three. The degree is the power of that highest derivative once the equation is polynomial in its derivatives. The third derivative appears only to the first power, so the degree is one. The fourth power sits on the second derivative, a lower-order term, and degree is concerned only with the power of the highest-order derivative, not with any lower ones.
Solution 2. Because the right-hand side depends only on x, integrate term by term.
Solution 3. Substitute x equals 0 and y equals 3 into the general solution. Every term with x vanishes, leaving 3 equals the constant, so the constant is 3.
Solution 4. Integrate once to obtain the first derivative.
Applying the condition that the first derivative equals 5 at x equals 0 gives the first constant as 5. Integrate again to obtain the function.
Applying the condition that the function equals minus 1 at x equals 0 gives the second constant as minus 1.
Solution 5. The equation is separable. Divide by y and multiply by dx to gather each variable on its own side.
Integrating both sides gives the natural logarithm of y equal to x squared plus a constant, and exponentiating turns it into a clean exponential form.
Applying the condition that y equals 3 when x equals 0, where the exponential is 1, gives the constant as 3.
Solution 6. The equation is linear with the coefficient of y equal to one over x, so the integrating factor is the exponential of the integral of one over x, which is the exponential of the natural logarithm of x, namely x itself.
Multiplying the equation by x makes the left side the derivative of a product.
Integrating gives x times y equal to x cubed plus a constant, and dividing by x gives the general solution.
Solution 7. Substituting an exponential trial gives the auxiliary equation, which factors directly.
The roots are minus 3 and 2, two distinct real values, so the general solution is a combination of two exponentials.
Solution 8. The auxiliary equation does not factor over the reals, so apply the quadratic formula.
The real part is 2 and the imaginary part is 3, so the solution is an exponential envelope multiplying a sine and cosine at frequency 3.
Solution 9. First find the complementary function by solving the homogeneous auxiliary equation.
The roots are 2 and minus 1, giving the complementary function.
Because the right-hand side is a degree-one polynomial, try a general degree-one polynomial for the particular integral, whose first derivative is a constant and whose second derivative is zero. Substituting into the equation and collecting terms gives the following match.
Matching the coefficient of x gives alpha equal to minus 2, and matching the constant term then gives beta equal to 1, so the particular integral is the line below.
The general solution is the sum of the two parts.
Solution 10. The excess demand is demand minus supply, which simplifies to a single linear expression in price.
Setting the rate of change to zero gives the equilibrium price, since 90 minus 5 times the equilibrium price is zero, so the equilibrium price is 18. Rewriting around that equilibrium shows the price is pulled toward it.
This is separable. Separating and integrating gives the natural logarithm of the gap to equilibrium as a linear function of time, and exponentiating produces a decaying exponential.
Applying the starting price of 5 gives the constant as minus 13.
As time grows, the exponential term decays to zero, so the price converges to the long-run equilibrium of 18.
See you soon.
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