Practice Flashcards
When is a first-order ODE separable?
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All Flashcards in Topic 5.18
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5.18.18 cards
When is a first-order ODE separable?
When dy/dx can be written as f(x)·g(y) — an x-part times a y-part.
How do you separate the variables?
Divide by g(y), multiply by dx: collect all y's with dy on the left, all x's with dx on the right, then integrate.
How many constants of integration after separating and integrating?
Just one — put a single +C on the right-hand side.
What is an initial condition used for?
To find the constant C: substitute the known point, giving the one particular solution through it.
Solve dy/dx = xy (general solution).
(1/y)dy = x dx ⇒ ln|y| = x²/2 + C ⇒ y = A e^(x²/2).
∫(1/(y − a)) dy = ?
ln|y − a| + C.
dT/dt = −k(T − r): what is the limiting temperature as t → ∞?
T → r (room temperature), since the exponential term decays to 0.
Why should you substitute the initial condition before rearranging?
The algebra is usually simpler in the un-rearranged form (e.g. ln form), reducing errors.
5.18.28 cards
What is the integrating factor for dy/dx + P(x)y = Q(x)?
I = e^(∫P dx). Multiply through by it; the left side becomes (Iy)′.
After multiplying a linear ODE by its integrating factor I, what is the left side?
Exactly (I·y)′ — so integrate to get Iy = ∫IQ dx.
How do you spot P(x) in dy/dx + P(x)y = Q(x)?
Write the equation with dy/dx alone (coefficient 1); P(x) is then the coefficient of y.
When is a first-order ODE homogeneous?
When dy/dx can be written using only the ratio y/x (e.g. (x+y)/x = 1 + y/x).
What substitution solves a homogeneous ODE, and what is dy/dx then?
Let y = vx; then dy/dx = v + x dv/dx (product rule). It becomes separable in v and x.
After solving in v, what's the final step?
Replace v by y/x, then apply the initial condition.
∫cot x dx = ?
ln|sin x| + C (so e^(∫cot x dx) = sin x).
Integrating factor for dy/dx + (2/x)y = x (x > 0)?
∫(2/x)dx = ln x², so I = x²; then (x²y)′ = x³.
5.18.38 cards
State Euler's method for dy/dx = f(x, y).
x_(n+1) = x_n + h, y_(n+1) = y_n + h·f(x_n, y_n), where h is the step size.
What does Euler's method actually do geometrically?
Follows the tangent (gradient f) in small straight steps of length h, approximating the curve by line segments.
Why is Euler's method only an approximation?
It uses the gradient at the START of each step, so it drifts off a curving solution; error grows with step size h.
If the solution curve is concave up, does Euler over- or under-estimate?
Underestimate — the tangents lie below a concave-up curve.
If the solution curve is concave down, does Euler over- or under-estimate?
Overestimate — the tangents lie above a concave-down curve.
How do you find the error of an Euler estimate?
error = |y_exact − y_Euler|, when the exact solution is known.
What happens to the error if you halve the step size h?
It roughly halves (error ∝ h), but you need twice as many steps.
Which paper most often features Euler's method?
Paper 3 (the extended-investigation paper), though it also appears in Paper 2.
Topic 5.18 study notes
Full notes & explanations for Differential equations (HL only)
Math AA exam skills
Paper structures, command terms & tips
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