Use Euler’s method: \(y_{n+1} = y_n + h \cdot f(x_n, y_n)\), where \(f(x, y) = \frac{4x^2 + y^2 – xy}{x^2}\), \(h = 0.1\), and initial condition \(y(1) = 2\).

Step 1: At \(x_0 = 1\), \(y_0 = 2\):

\[f(1, 2) = \frac{4(1)^2 + (2)^2 – (1)(2)}{(1)^2} = \frac{4 + 4 – 2}{1} = 6\]

\[y_1 = 2 + 0.1 \cdot 6 = 2.6 \quad (x_1 = 1.1)\]

Step 2: At \(x_1 = 1.1\), \(y_1 = 2.6\):

\[f(1.1, 2.6) = \frac{4(1.1)^2 + (2.6)^2 – (1.1)(2.6)}{(1.1)^2} = \frac{4(1.21) + 6.76 – 2.86}{1.21} = \frac{4.84 + 6.76 – 2.86}{1.21} = \frac{8.74}{1.21} \approx 7.22\]

\[y_2 = 2.6 + 0.1 \cdot 7.22 = 2.6 + 0.722 = 3.322 \quad (x_2 = 1.2)\]

Step 3: At \(x_2 = 1.2\), \(y_2 = 3.322\):

\[f(1.2, 3.322) = \frac{4(1.2)^2 + (3.322)^2 – (1.2)(3.322)}{(1.2)^2} = \frac{4(1.44) + 11.035684 – 3.9864}{1.44} \approx \frac{5.76 + 11.035684 – 3.9864}{1.44} \approx 8.86\]

\[y_3 = 3.322 + 0.1 \cdot 8.86 = 3.322 + 0.886 = 4.208 \quad (x_3 = 1.3)\]

Step 4: At \(x_3 = 1.3\), \(y_3 = 4.208\):

\[f(1.3, 4.208) = \frac{4(1.3)^2 + (4.208)^2 – (1.3)(4.208)}{(1.3)^2} = \frac{4(1.69) + 17.707264 – 5.4704}{1.69} \approx \frac{6.76 + 17.707264 – 5.4704}{1.69} \approx 11.26\]

\[y_4 = 4.208 + 0.1 \cdot 11.26 = 4.208 + 1.126 = 5.334 \quad (x_4 = 1.4)\]

Thus, \(y(1.4) \approx 5.34\).

Solve for isoclines where \(\frac{dy}{dx} = 4\):

\[\frac{4x^2 + y^2 – xy}{x^2} = 4\]

\[y^2 – xy = 0\]

\[y(y – x) = 0\]

Thus, \(y = 0\) or \(y = x\).

Sketch shows two straight lines: the x-axis (\(y = 0\)) and the line \(y = x\).

Complete the square for \(m^2 – 2m + 4\):

\[m^2 – 2m + 4 = (m^2 – 2m + 1) + 3 = (m – 1)^2 + 3\]

Thus, \(a = 1\), \(b = 3\).

Recognize the differential equation as homogeneous:

\[\frac{dy}{dx} = \frac{4x^2 + y^2 – xy}{x^2} = 4 + \left(\frac{y}{x}\right)^2 – \frac{y}{x}\]

Let \(y = vx\), so \(\frac{dy}{dx} = v + x \frac{dv}{dx}\):

\[v + x \frac{dv}{dx} = 4 + v^2 – v\]

\[x \frac{dv}{dx} = v^2 – 2v + 4\]

Separate variables: \[\int \frac{1}{v^2 – 2v + 4} \, dv = \int \frac{1}{x} \, dx\]

From part (c)(i), \(v^2 – 2v + 4 = (v – 1)^2 + 3\):

\[\int \frac{1}{(v – 1)^2 + 3} \, dv = \int \frac{1}{x} \, dx\]

Left side: \[\frac{1}{\sqrt{3}} \arctan\left(\frac{v – 1}{\sqrt{3}}\right) + C_1\]

Right side: \[\ln x + C_2\]

\[\frac{1}{\sqrt{3}} \arctan\left(\frac{v – 1}{\sqrt{3}}\right) = \ln x + c\]

Use initial condition \(x = 1\), \(y = 2\), so \(v = \frac{2}{1} = 2\):

\[\frac{1}{\sqrt{3}} \arctan\left(\frac{2 – 1}{\sqrt{3}}\right) = \ln 1 + c\]

\[\frac{1}{\sqrt{3}} \arctan\left(\frac{1}{\sqrt{3}}\right) = c\]

\[\arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}\], so \(c = \frac{\pi}{6 \sqrt{3}}\)

\[\frac{1}{\sqrt{3}} \arctan\left(\frac{v – 1}{\sqrt{3}}\right) = \ln x + \frac{\pi}{6 \sqrt{3}}\]

\[\arctan\left(\frac{v – 1}{\sqrt{3}}\right) = \sqrt{3} \ln x + \frac{\pi}{6}\]

Substitute \(v = \frac{y}{x}\):

\[\frac{y}{x} – 1 = \sqrt{3} \tan\left(\sqrt{3} \ln x + \frac{\pi}{6}\right)\]

\[y = x \left( \sqrt{3} \tan\left(\sqrt{3} \ln x + \frac{\pi}{6}\right) + 1 \right)\]

Sketch the graph of \(y = x \left( \sqrt{3} \tan\left(\sqrt{3} \ln x + \frac{\pi}{6}\right) + 1 \right)\) for \(1 \leq x \leq 1.4\).

Curve drawn over the correct domain, passing through \((1, 2)\) and increasing.

The sketch in part (c)(iii) shows that \(f\) is concave up (or \(f’\) is increasing).

This implies that Euler’s method, which uses tangent lines, will underestimate the true value of \(f(x)\). Thus, \(f(1.4) > 5.34\).

(a)

(b)

(c)(i)

(c)(ii)

(c)(iii)

(c)(iv)