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.

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.

Discuss in Forum

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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