Questions on series solutions of differential equations

Question 1

Describe the Taylor series method for solving a differential equation with given initial conditions.

Accepted Answer

Rearrange for the highest derivative and substitute the initial values to find it at the starting point. Differentiate the whole equation and substitute again for the next derivative, and so on. Each derivative becomes one coefficient of the Taylor series of the solution. B1 for evaluating the highest derivative from the initial values, B1 for repeated differentiation supplying each further coefficient.

Question 2

For dy/dx = x + y² with y = 1 at x = 0, find the values of the first and second derivatives of y at x = 0.

Accepted Answer

dy/dx at 0 is 0 + 1² = 1. Differentiating the equation: the second derivative is 1 + 2y(dy/dx), which at 0 is 1 + 2(1)(1) = 3. B1 for dy/dx = 1, M1 for differentiating the equation, A1 for 3.

Question 3

For d²y/dx² = e^y with y = 0 and dy/dx = 1 at x = 0, find the values of the second and third derivatives of y at x = 0.

Accepted Answer

At the origin y″ = e⁰ = 1. Differentiating the equation with the chain rule, y‴ = e^yy′, which at the origin is 1 × 1 = 1. B1 for y″(0), B1 for y‴(0). The equation is differentiated as it stands; e^y is never expanded, only evaluated.

Question 4

The substitution z = 1/y is to be used on dy/dx + y = xy². State what the substitution assumes about y, and write down the solution of the equation that it therefore cannot produce.

Accepted Answer

z = 1/y requires y ≠ 0. B1. The constant function y = 0 satisfies the equation, since 0 + 0 = x × 0, but under the substitution it corresponds to no finite z, so it has to be listed separately as y = 0 beside the family the substitution finds. B1. Dividing by y², which is what the substitution amounts to, is where that solution is removed.

Question 5

For dy/dx = x + y² with y = 1 at x = 0, the first and second derivatives of y at x = 0 are 1 and 3. Find the third derivative at x = 0 and write the series solution up to the term in x³.

Accepted Answer

The second derivative is 1 + 2y y′. Differentiating that by the product rule gives 2(y′)² + 2y y″, which at the origin is 2(1)² + 2(1)(3) = 8.
Taylor's series about 0 is y = y(0) + y′(0)x + y″(0)x²/2! + y‴(0)x³/3!, so y = 1 + x + 3x²/2! + 8x³/3!, that is y = 1 + x + 1.5x² + (4/3)x³ + …. M1 for differentiating by the product rule, A1 for 8, M1 for the Taylor form with its factorials, A1 for the series. Keep the factorials visible until the last line; the divisor is n!, not n.

Question 6

For dy/dx = x + e^y with y = 0 at x = 0, find the values of the first three derivatives of y at x = 0, and hence write the series solution up to and including the term in x³.

Accepted Answer

At the origin, dy/dx = 0 + e⁰ = 1.
Differentiate the equation, using the chain rule on e^y: y″ = 1 + e^yy′, which at the origin is 1 + (1)(1) = 2.
Differentiate again, by the product rule on e^yy′: y‴ = e^y(y′)² + e^yy″, which at the origin is (1)(1)² + (1)(2) = 3.
Taylor's series about 0 is y = y(0) + y′(0)x + y″(0)x²/2! + y‴(0)x³/3!, so y = 0 + x + 2x²/2! + 3x³/3!, that is y = x + x² + x³/2 + …. B1 for dy/dx = 1 at the origin, M1 for differentiating the equation with the chain rule on e^y, A1 for 2 and 3, A1 for the series.

Question 7

For a second order equation, explain why two initial conditions are needed before the Taylor series method can begin.

Accepted Answer

The equation delivers the second derivative in terms of x, y and dy/dx, so both y and dy/dx must be known at the starting point before anything can be evaluated. With only one condition the very first substitution is impossible. B1 B1 for the two points: the equation returns the second derivative in terms of y and dy/dx, and both must be known before any substitution.

Question 8

The function y satisfies dy/dx = y² − x, and y = 2 when x = 1. Find the second and third derivatives of y at x = 1, and hence obtain the Taylor series solution in ascending powers of (x − 1) up to and including the term in (x − 1)³.

Accepted Answer

At x = 1 the equation gives y′ = 2² − 1 = 3.
Differentiate the equation: y″ = 2y y′ − 1, so at x = 1 this is 2(2)(3) − 1 = 11.
Differentiate again, by the product rule on 2y y′: y‴ = 2(y′)² + 2y y″, so at x = 1 this is 2(3)² + 2(2)(11) = 18 + 44 = 62.
Taylor's series about x = 1 runs in powers of (x − 1):
y = y(1) + y′(1)(x − 1) + y″(1)(x − 1)²/2! + y‴(1)(x − 1)³/3!
= 2 + 3(x − 1) + 5.5(x − 1)² + (31/3)(x − 1)³ + …
M1 for differentiating the equation, A1 for y″(1) = 11, M1 for the product rule on the next differentiation, A1 for y‴(1) = 62, A1 for the series in powers of (x − 1). The information is given at x = 1, not at the origin, so a Maclaurin series is not available and every bracket must be (x − 1).

Question 9

The function y satisfies dy/dx = y − x, and y = 2 when x = 0. Find the Maclaurin series solution up to and including the term in x⁴. Verify that y = 1 + x + e^x satisfies both the differential equation and the condition y = 2 when x = 0, and, taking it as the exact solution, compare the accuracy of your series at x = 0.2 with its accuracy at x = 2.

Accepted Answer

At the origin y′ = 2 − 0 = 2. Differentiating the equation gives y″ = y′ − 1, so y″(0) = 1, and every further differentiation gives y‴ = y″, y⁗ = y‴, so y‴(0) = y⁗(0) = 1.
Hence y = 2 + 2x + x²/2! + x³/3! + x⁴/4! = 2 + 2x + x²/2 + x³/6 + x⁴/24.
Checking the exact solution: if y = 1 + x + e^x then dy/dx = 1 + e^x, while y − x = 1 + e^x, and y(0) = 2 as required.
At x = 0.2 the series gives 2.42140 and the exact value is 2.421403, an error of about 3 × 10⁻⁶. At x = 2 the series gives 10 while the exact value is 10.389, an error of about 0.39.
M1 for the repeated differentiation, A1 for the derivatives 2, 1, 1, 1 at the origin, A1 for the series, B1 for verifying that the given function satisfies the equation and the condition, B1 for the comparison with figures. A truncated series is a local statement. The neglected terms carry powers of x, so the further from the expansion point the estimate is taken, the more the tail matters, and doubling the distance does far more than double the error.

Question 10

Use the substitution z = y − x to solve the differential equation dy/dx = (y − x)² + 1, giving y in terms of x, and state any solution that the method does not produce.

Accepted Answer

With z = y − x, dz/dx = dy/dx − 1 = (y − x)² + 1 − 1 = z². M1 for transforming the derivative, A1 for dz/dx = z².
Separating, dz/z² = dx, and integrating gives −1/z = x + c, so z = −1/(x + c). M1 for separating and integrating.
Translating back, y = x − 1/(x + c). A1.
Dividing by z² assumed z ≠ 0, and z = 0 satisfies dz/dx = z² on its own, so y = x is also a solution and is not a member of the family for any finite c. B1 for listing it. Check: y = x gives dy/dx = 1 and (y − x)² + 1 = 1. The substitution turned a nonlinear equation in y into a separable one in z because the right side depended on y only through y − x.

Question 11

Given that d²y/dx² = y dy/dx + x with y = 1 and dy/dx = 1 at x = 0, find the values of the second and third derivatives at x = 0, and write the series solution up to and including the term in x³.

Accepted Answer

At the origin y″ = (1)(1) + 0 = 1. B1.
Differentiate the equation, using the product rule on yy′: y‴ = (y′)² + yy″ + 1, which at the origin is 1 + 1 + 1 = 3. M1 for the product rule, A1.
Taylor's series about 0: y = 1 + x + x²/2! + 3x³/3! = 1 + x + x²/2 + x³/2 + …. A1. Differentiating yy′ as y′ alone, or as y′y″, changes the third derivative and every coefficient after it; the product rule gives two terms and both are needed.

Question 12

Given that y'' + xy' + 2y = 0 with y = 1 and dy/dx = 0 at x = 0, find the series solution for y up to and including the term in x⁶. Hence estimate the value of y at x = 0.2, giving your answer to five decimal places.

Accepted Answer

Rearrange for the highest derivative: y″ = −xy′ − 2y. At the origin y″ = −0 − 2(1) = −2.
Differentiate, using the product rule on xy′:
y‴ = −y′ − xy″ − 2y′ = −3y′ − xy″, so y‴(0) = 0.
Again: y⁗ = −3y″ − y″ − xy‴ = −4y″ − xy‴, so y⁗(0) = −4(−2) = 8.
The pattern holds, each differentiation raising the coefficient by one. The fifth derivative is −5y‴ − xy⁗, giving 0, and the sixth is −6y⁗ − xy⁽⁵⁾, giving −6(8) = −48.
Assembling Taylor's series:
y = 1 + 0 − 2x²/2! + 0 + 8x⁴/4! + 0 − 48x⁶/6!
= 1 − x² + x⁴/3 − x⁶/15.
Every odd derivative vanishes, because each is a multiple of the one two places before it and the chain starts at y′(0) = 0.
At x = 0.2: y ≈ 1 − 0.04 + 0.0016/3 − 0.000064/15 = 1 − 0.04 + 0.00053333 − 0.00000427 = 0.96053.
M1 for rearranging for y″, A1 for y″(0) = −2, M1 for the repeated differentiation, A1 for y⁗(0) = 8, A1 for −48 at the sixth derivative, A1 for the assembled series, B1 for 0.96053. The terms shrink by roughly a factor of a hundred each time, so the next one cannot disturb the fifth decimal place.

Question 13

Given that y = tan x, show that dy/dx = 1 + y². Hence, by repeated differentiation, find the Maclaurin series for tan x up to and including the term in x⁵.

Accepted Answer

Differentiating, dy/dx = sec²x = 1 + tan²x = 1 + y², using the identity sec²x = 1 + tan²x.
At x = 0, y = tan 0 = 0 and y′ = 1 + 0 = 1.
Differentiate the equation each time rather than the function:
y″ = 2y y′, so y″(0) = 0.
y‴ = 2(y′)² + 2y y″, so y‴(0) = 2(1)² = 2.
Fourth derivative: 6y′y″ + 2y y‴, which at 0 is 0.
Fifth derivative: 6(y″)² + 8y′y‴ + 2y y⁗, which at 0 is 8(1)(2) = 16.
Assembling Taylor's series about the origin:
tan x = 0 + x + 0 + 2x³/3! + 0 + 16x⁵/5!
= x + x³/3 + 2x⁵/15.
B1 for dy/dx = 1 + y², M1 for differentiating the equation to reach y″ = 2y y′, A1 for y‴(0) = 2, M1 for continuing to the fifth derivative, A1 for 16, M1 for the Taylor form with its factorials, A1 for the printed series.
Differentiating 1 + y² is far lighter than differentiating sec²x five times. And tan is odd, so every even derivative at the origin vanishes; noticing that after y″(0) = 0 saves checking the fourth one. As a test, the series gives 0.309324 at x = 0.3 against tan 0.3 = 0.309336.

Question 14

Use the substitution x = e^u, for x > 0, to show that the equation x² d²y/dx² + x dy/dx − 4y = 0 becomes d²y/du² − 4y = 0. Solve the transformed equation, and hence write down the general solution of the original equation in terms of x.

Accepted Answer

With x = e^u, du/dx = 1/x, so dy/dx = (dy/du)(du/dx) = (1/x) dy/du. M1 for the chain rule.
Differentiating again with the product rule: d²y/dx² = −(1/x²) dy/du + (1/x) d/dx(dy/du) = −(1/x²) dy/du + (1/x²) d²y/du². A1 for the second derivative.
Substituting: x² d²y/dx² = d²y/du² − dy/du and x dy/dx = dy/du, so the equation becomes d²y/du² − dy/du + dy/du − 4y = 0, that is d²y/du² − 4y = 0. M1 for substituting both derivatives, A1.
The auxiliary equation m² − 4 = 0 has roots ±2, so y = Ae^2u + Be^−2u. M1.
Since e^u = x, e^2u = x² and e^−2u = 1/x², giving y = Ax² + B/x². A1. The substitution is chosen because the equation has x² against the second derivative and x against the first, and multiplying each derivative by the matching power of x is exactly what differentiating with respect to ln x does. The restriction x > 0 is what makes u = ln x exist; for x < 0 the same working with x = −e^u returns the same family.

Question 15

Use the substitution y = zx to solve dy/dx = (x² + y²)/(xy) for x > 0, giving y² in terms of x.

Accepted Answer

With y = zx, dy/dx = z + x dz/dx by the product rule, and the right side is (x² + z²x²)/(x · zx) = (1 + z²)/z. M1 for transforming both sides.
So z + x dz/dx = (1 + z²)/z, and subtracting z leaves x dz/dx = 1/z. A1.
Separating, z dz = dx/x, and integrating gives z²/2 = ln x + c. M1 for separating, A1.
With z = y/x: y²/(2x²) = ln x + c, so y² = x²(2 ln x + C), where C = 2c. A1. The right side of the original equation depends on x and y only through the ratio y/x, which is the signal that y = zx will separate it. The division by xy assumes y ≠ 0, but y = 0 is not a solution of the original equation, since the right side is then undefined, so nothing has been lost.