Numerical Solution of ODEs Questions and Answers

Q: Using Euler's method taking step size = 0.1, the approximate value of \(y\) obtained corresponding to \(x = 0.2\) for the initial value problem \(\frac{dy}{dx} = x^{2} + y^{2}\) and \(y(0) = 1\), is

Euler's method gives 1.222.Euler's method updates \(y_{n+1} = y_{n} + h\,f(x_{n}, y_{n})\) with \(f(x, y) = x^{2} + y^{2}\) and \(h = 0.1\).\[ \begin{aligned} y_{1} &= 1 + 0.1(0^{2} + 1^{2}) = 1.1 \\ y_{2} &= 1.1 + 0.1(0.1^{2} + 1.1^{2}) \\ &= 1.1 + 0.1(0.01 + 1.21) = 1.222 \end{aligned} \]So \(y(0.2) \approx 1.222\). The value 1.122 adds the second increment, 0.122, to the starting value 1 instead of to 1.1; 1.110 and 1.322 do not come from the two Euler steps.

Q: For the initial value problem\(y' = f(x, y),\ y(x_{0}) = y_{0}\),generate approximations \(y_{n}\) to \(y(x_{n})\), \(x_{n} = x_{0} + nh\), for a fixed \(h \gt 0\) and \(n = 1, 2, 3, \ldots\), using the recursion formula\(y_{n} = y_{n-1} + ak_{1} + bk_{2}\), where\(k_{1} = hf(x_{n-1}, y_{n-1})\) and \(k_{2} = hf(x_{n-1} + \alpha h, y_{n-1} + \beta k_{1})\).Which one of the following choices of \(a, b, \alpha, \beta\) for the above recursion formula gives the Runge-Kutta method of order 2 ?

The second-order Runge-Kutta choice is \(a = 0.25, b = 0.75, \alpha = 2/3, \beta = 2/3\).Matching the Taylor expansion of \(y(x_{n})\) to second order requires\[ \begin{aligned} a + b &= 1 \\ b\alpha &= \tfrac{1}{2},\quad b\beta = \tfrac{1}{2} \end{aligned} \]For \(a = 0.25, b = 0.75, \alpha = \beta = 2/3\): \(a + b = 1\) and \(b\alpha = b\beta = 0.5\). This is Ralston's method.\(a = b = 1\) gives \(a + b = 2\). \(a = b = 0.5, \alpha = \beta = 2\) gives \(b\alpha = 1\). \(a = b = 0.5, \alpha = 1, \beta = 2\) gives \(b\beta = 1\). Each fails a condition.

Q: For a given initial value problem\[ y' = 1 + xy,\ y(0) = 2, \]the value of \(y(0.1)\) by Euler's method with \(n = 0.1\) is :

Euler's method gives \(y(0.1) = 2.1\).The printed \(n = 0.1\) is the step size \(h\). Euler's method uses \(y_{1} = y_{0} + h f(x_{0}, y_{0})\) with \(f(x, y) = 1 + xy\):\[ y_{1} = 2 + 0.1(1 + 0 \cdot 2) = 2 + 0.1 = 2.1 \]The slope at the start is only 1, because the \(xy\) term vanishes at \(x = 0\). 2.2 and 2.4 would need slopes of 2 and 4, and 2.0 ignores the step altogether.

Q: For a given initial value problem\(y' = y - x\), \(y(0) = 2\)the value of \(y(0.1)\) by Runge-Kutta second order method is :

\(y(0.1) \approx\) 2.2050.With \(f(x, y) = y - x\), \(h = 0.1\) and the second-order Runge-Kutta (Heun) method:\[ \begin{aligned} k_{1} &= hf(0, 2) = 0.1 \times 2 = 0.2 \\ k_{2} &= hf(0.1, 2.2) = 0.1 \times 2.1 = 0.21 \\ y_{1} &= 2 + \tfrac{1}{2}(0.2 + 0.21) = 2.205 \end{aligned} \]The midpoint form of RK2 gives the same value here. The exact solution \(y = x + 1 + e^{x}\) gives 2.2052, so the answer is accurate. 2.2100 adds \(k_{2}\) alone (an Euler step with the end slope), and 2.0050 and 2.1900 do not follow from the method.

Q: Let \(\dfrac{dy}{dx} = 1 + xy\) with \(y(0) = 2\). Using Picard's method, the first approximation is :

The first approximation is y1 = 2 + x + x2.Picard's iteration is \(y_{n+1}(x) = y_{0} + \int_{0}^{x} f(t, y_{n}(t))\, dt\), starting from \(y_{0} = 2\). With \(f(x, y) = 1 + xy\):\[ \begin{aligned} y_{1} &= 2 + \int_{0}^{x} (1 + 2t)\, dt \\ &= 2 + x + x^{2} \end{aligned} \]Leaving out the initial value 2 gives \(x + x^{2}\), and dropping the \(xy\) term gives \(2 + x\). The constant 20 has no source in the problem.

Q: Let \(\dfrac{dy}{dx} = f(x, y)\) be an initial value problem with \(y(x_{0}) = y_{0}\). Then in Runge-Kutta method of order 4, the value of K3 to find first approximation is :

The third slope is K3 = h f(x0 + h/2, y0 + k2/2).The classical fourth order Runge-Kutta method uses\[ \begin{aligned} k_{1} &= hf(x_{0}, y_{0}) \\ k_{2} &= hf(x_{0} + \tfrac{h}{2}, y_{0} + \tfrac{k_{1}}{2}) \\ k_{3} &= hf(x_{0} + \tfrac{h}{2}, y_{0} + \tfrac{k_{2}}{2}) \\ k_{4} &= hf(x_{0} + h, y_{0} + k_{3}) \end{aligned} \]and \(y_{1} = y_{0} + \frac{1}{6}(k_{1} + 2k_{2} + 2k_{3} + k_{4})\). The option \(hf(x_{0}, y_{0})\) is \(k_{1}\), and the step to \(x_{0} + h\) belongs to \(k_{4}\) (with \(k_{3}\), not h, added to y). The ratio \(k_{1}/k_{2}\) appears nowhere in the method.

Q: A reservoir discharges water through holes with shutters at depth h and surface area A(h). The values are given in the following table. If t denotes the time in minutes and the rate of fall of the water level is given by the equation \(\frac{dh}{dt} = -\frac{48h}{A}\), what is the time taken in minutes for the water level to fall from 14 ft. to 10 ft. above the holes ?h (in ft.)A (in sq. ft.)10940111078121200131352141540

The time taken is about 8.41 minutes.Separate the variables: \(dt = -\frac{A}{48h}\,dh\), so the time to fall from 14 ft to 10 ft is \(t = \int_{10}^{14}\frac{A}{48h}\,dh\). Tabulate \(f = \frac{A}{48h}\):\(f(10) = 1.9583\), \(f(11) = 2.0417\), \(f(12) = 2.0833\), \(f(13) = 2.1667\), \(f(14) = 2.2917\).Simpson's 1/3 rule with step 1:\[ \begin{aligned} t &= \tfrac{1}{3}[f_{0} + 4(f_{1} + f_{3}) + 2f_{2} + f_{4}] \\ &= \tfrac{1}{3}[1.9583 + 16.8333 + 4.1667 + 2.2917] \\ &= \tfrac{1}{3}(25.25) \approx 8.42 \end{aligned} \]which is the keyed 8.41. The other options are far off: the integrand lies between about 1.96 and 2.29 over a 4 ft fall, so the time must lie between about 7.8 and 9.2 minutes.

Using Euler's method taking step size = 0.1, the approximate value of \(y\) obtained corresponding to \(x = 0.2\) for the initial value problem \(\frac{dy}{dx} = x^{2} + y^{2}\) and \(y(0) = 1\), is

A.1.322
B.1.122
C.1.222
D.1.110

Exam Year(s)

GATE Mathematics - 2012,

Answer: C

Euler's method gives 1.222.

Euler's method updates \(y_{n+1} = y_{n} + h\,f(x_{n}, y_{n})\) with \(f(x, y) = x^{2} + y^{2}\) and \(h = 0.1\).

\[ \begin{aligned} y_{1} &= 1 + 0.1(0^{2} + 1^{2}) = 1.1 \\ y_{2} &= 1.1 + 0.1(0.1^{2} + 1.1^{2}) \\ &= 1.1 + 0.1(0.01 + 1.21) = 1.222 \end{aligned} \]

So \(y(0.2) \approx 1.222\). The value 1.122 adds the second increment, 0.122, to the starting value 1 instead of to 1.1; 1.110 and 1.322 do not come from the two Euler steps.

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For the initial value problem

\(y' = f(x, y),\ y(x_{0}) = y_{0}\),

generate approximations \(y_{n}\) to \(y(x_{n})\), \(x_{n} = x_{0} + nh\), for a fixed \(h \gt 0\) and \(n = 1, 2, 3, \ldots\), using the recursion formula

\(y_{n} = y_{n-1} + ak_{1} + bk_{2}\), where

\(k_{1} = hf(x_{n-1}, y_{n-1})\) and \(k_{2} = hf(x_{n-1} + \alpha h, y_{n-1} + \beta k_{1})\).

Which one of the following choices of \(a, b, \alpha, \beta\) for the above recursion formula gives the Runge-Kutta method of order 2 ?

A.\(a = 1, b = 1, \alpha = 0.5, \beta = 0.5\)
B.\(a = 0.5, b = 0.5, \alpha = 2, \beta = 2\)
C.\(a = 0.25, b = 0.75, \alpha = 2/3, \beta = 2/3\)
D.\(a = 0.5, b = 0.5, \alpha = 1, \beta = 2\)

Exam Year(s)

GATE Mathematics - 2024,

Answer: C

The second-order Runge-Kutta choice is \(a = 0.25, b = 0.75, \alpha = 2/3, \beta = 2/3\).

Matching the Taylor expansion of \(y(x_{n})\) to second order requires

\[ \begin{aligned} a + b &= 1 \\ b\alpha &= \tfrac{1}{2},\quad b\beta = \tfrac{1}{2} \end{aligned} \]

For \(a = 0.25, b = 0.75, \alpha = \beta = 2/3\): \(a + b = 1\) and \(b\alpha = b\beta = 0.5\). This is Ralston's method.

\(a = b = 1\) gives \(a + b = 2\). \(a = b = 0.5, \alpha = \beta = 2\) gives \(b\alpha = 1\). \(a = b = 0.5, \alpha = 1, \beta = 2\) gives \(b\beta = 1\). Each fails a condition.

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For a given initial value problem

\[ y' = 1 + xy,\ y(0) = 2, \]

the value of \(y(0.1)\) by Euler's method with \(n = 0.1\) is :

A.2.1
B.2.0
C.2.2
D.2.4

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

Euler's method gives \(y(0.1) = 2.1\).

The printed \(n = 0.1\) is the step size \(h\). Euler's method uses \(y_{1} = y_{0} + h f(x_{0}, y_{0})\) with \(f(x, y) = 1 + xy\):

\[ y_{1} = 2 + 0.1(1 + 0 \cdot 2) = 2 + 0.1 = 2.1 \]

The slope at the start is only 1, because the \(xy\) term vanishes at \(x = 0\). 2.2 and 2.4 would need slopes of 2 and 4, and 2.0 ignores the step altogether.

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For a given initial value problem

\(y' = y - x\), \(y(0) = 2\)

the value of \(y(0.1)\) by Runge-Kutta second order method is :

A.2.2100
B.2.0050
C.2.2050
D.2.1900

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

\(y(0.1) \approx\) 2.2050.

With \(f(x, y) = y - x\), \(h = 0.1\) and the second-order Runge-Kutta (Heun) method:

\[ \begin{aligned} k_{1} &= hf(0, 2) = 0.1 \times 2 = 0.2 \\ k_{2} &= hf(0.1, 2.2) = 0.1 \times 2.1 = 0.21 \\ y_{1} &= 2 + \tfrac{1}{2}(0.2 + 0.21) = 2.205 \end{aligned} \]

The midpoint form of RK2 gives the same value here. The exact solution \(y = x + 1 + e^{x}\) gives 2.2052, so the answer is accurate. 2.2100 adds \(k_{2}\) alone (an Euler step with the end slope), and 2.0050 and 2.1900 do not follow from the method.

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Let \(\dfrac{dy}{dx} = 1 + xy\) with \(y(0) = 2\). Using Picard's method, the first approximation is :

A.\(y_{1} = x + x^{2}\)
B.\(y_{1} = 2 + x + x^{2}\)
C.\(y_{1} = 20 + x + x^{2}\)
D.\(y_{1} = 2 + x\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The first approximation is y₁ = 2 + x + x².

Picard's iteration is \(y_{n+1}(x) = y_{0} + \int_{0}^{x} f(t, y_{n}(t))\, dt\), starting from \(y_{0} = 2\). With \(f(x, y) = 1 + xy\):

\[ \begin{aligned} y_{1} &= 2 + \int_{0}^{x} (1 + 2t)\, dt \\ &= 2 + x + x^{2} \end{aligned} \]

Leaving out the initial value 2 gives \(x + x^{2}\), and dropping the \(xy\) term gives \(2 + x\). The constant 20 has no source in the problem.

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Let \(\dfrac{dy}{dx} = f(x, y)\) be an initial value problem with \(y(x_{0}) = y_{0}\). Then in Runge-Kutta method of order 4, the value of K₃ to find first approximation is :

A.\(K_{3} = h(f(x_{0}, y_{0}))\)
B.\(K_{3} = h(f(x_{0} + h, y_{0} + h))\)
C.\(K_{3}=h\left(f\left(x_{0}+\frac{h}{2},y_{0}+\frac{k_{2}}{2}\right)\right)\)
D.\(K_{3}=h\left(f\left(x_{0}+\frac{h}{2},y_{0}+\frac{k_{1}}{k_{2}}\right)\right)\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The third slope is K₃ = h f(x₀ + h/2, y₀ + k₂/2).

The classical fourth order Runge-Kutta method uses

\[ \begin{aligned} k_{1} &= hf(x_{0}, y_{0}) \\ k_{2} &= hf(x_{0} + \tfrac{h}{2}, y_{0} + \tfrac{k_{1}}{2}) \\ k_{3} &= hf(x_{0} + \tfrac{h}{2}, y_{0} + \tfrac{k_{2}}{2}) \\ k_{4} &= hf(x_{0} + h, y_{0} + k_{3}) \end{aligned} \]

and \(y_{1} = y_{0} + \frac{1}{6}(k_{1} + 2k_{2} + 2k_{3} + k_{4})\). The option \(hf(x_{0}, y_{0})\) is \(k_{1}\), and the step to \(x_{0} + h\) belongs to \(k_{4}\) (with \(k_{3}\), not h, added to y). The ratio \(k_{1}/k_{2}\) appears nowhere in the method.

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A reservoir discharges water through holes with shutters at depth h and surface area A(h). The values are given in the following table. If t denotes the time in minutes and the rate of fall of the water level is given by the equation \(\frac{dh}{dt} = -\frac{48h}{A}\), what is the time taken in minutes for the water level to fall from 14 ft. to 10 ft. above the holes ?

h (in ft.)	A (in sq. ft.)
10	940
11	1078
12	1200
13	1352
14	1540

A.8.41
B.2.63
C.5.48
D.6.06

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The time taken is about 8.41 minutes.

Separate the variables: \(dt = -\frac{A}{48h}\,dh\), so the time to fall from 14 ft to 10 ft is \(t = \int_{10}^{14}\frac{A}{48h}\,dh\). Tabulate \(f = \frac{A}{48h}\):

\(f(10) = 1.9583\), \(f(11) = 2.0417\), \(f(12) = 2.0833\), \(f(13) = 2.1667\), \(f(14) = 2.2917\).

Simpson's 1/3 rule with step 1:

\[ \begin{aligned} t &= \tfrac{1}{3}[f_{0} + 4(f_{1} + f_{3}) + 2f_{2} + f_{4}] \\ &= \tfrac{1}{3}[1.9583 + 16.8333 + 4.1667 + 2.2917] \\ &= \tfrac{1}{3}(25.25) \approx 8.42 \end{aligned} \]

which is the keyed 8.41. The other options are far off: the integrand lies between about 1.96 and 2.29 over a 4 ft fall, so the time must lie between about 7.8 and 9.2 minutes.

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