1.
Write the equation y' = 5 - 3y in the standard form y' + p(x)y = q(x). What is p(x)?
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Answer:
Move the 3y term to the left side.
So p(x) = 3 and q(x) = 5.
First-order linear practice problems
You solve equations that can be written as y' + p(x)y = q(x). The method has four steps. Put the equation in standard form. Build the integrating factor e^(∫p dx). Multiply both sides, integrate, and add C. Use the start value to fix C. Recognition cue: the equation contains y' and y to the first power only, never y^2, ln y, or y inside another function. You will practice: Standard form, Integrating factor, Full solution, Initial condition and long run.
2.
Find the integrating factor for y' + 4y = 7.
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Answer:
Here p(x) = 4. The factor is e raised to an antiderivative of p.
3.
Solve y' + 2y = 0 with y(0) = 3.
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Answer:
With q = 0 the solution is a constant times e^(-2x).
Check: y' = -6e^(-2x), which equals -2y.
4.
The current in a circuit obeys I' + 3I = 12, with t in seconds. For any starting current, what value does I(t) approach as t grows?
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Answer:
Every solution is the steady value plus a shrinking exponential.
The term Ce^(-3t) shrinks to 0, so I approaches 4.
5.
Find the general solution of y' + y = 2.
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Answer:
Multiply by the integrating factor e^x, then integrate.
6.
A cup of coffee starts at 90 °C in a room at 20 °C. Its temperature T(t), with t in minutes, obeys T' = -0.1(T - 20). Find T(t).
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Answer:
Standard form is T' + 0.1T = 2.
7.
Solve y' + 4y = 8 with y(0) = 5.
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Answer:
Solve in general, then fit C.
8.
The current I(t) in an RL circuit obeys I' + 2I = 10 with I(0) = 0, t in seconds. Find I(t).
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Answer:
The current climbs from 0 toward 5.
9.
A bacteria culture has mass m(t) grams after t hours, and m' = 0.5m + 2 with m(0) = 4. Find m(t).
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Answer:
Standard form is m' - 0.5m = 2, so p = -0.5.
10.
Find the general solution of y' + (1/x)y = 3 for x > 0.
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Answer:
The integral of 1/x is ln x, so the factor is x itself.
11.
Consider the initial value problem (t² − 9)y′ + t y = t², y(1) = 5. Do not solve it. State the largest open interval on which a unique solution is guaranteed to exist.
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Answer:
Divide by t² − 9 to get standard form. The coefficients are continuous except where t² − 9 = 0, so the breaks are at t = −3 and t = 3. The initial point t = 1 sits between them.
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