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Problem 5i) a) The curve is given by the equation (x^2-3) y + bx = 4. We need to show that (dy)/(dx) = (y(2x y + b))/(3-x^2). Step 1: Differentiate both sides of the equation with respect to x using implicit differentiation. We apply the product rule for the first term (d)/(dx)((x^2-3) y) = (2x) y + (x^2-3)((1)/(y)(dy)/(dx)). The derivative of bx is b. The derivative of 4 is 0. So, we have: 2x y + (x^2-3)(1)/(y)(dy)/(dx) + b = 0 Step 2: Isolate (dy)/(dx). Subtract 2x y and b from both sides: (x^2-3)(1)/(y)(dy)/(dx) = -2x y - b Multiply both sides by y: (x^2-3)(dy)/(dx) = -y(2x y + b) Divide both sides by (x^2-3): (dy)/(dx) = (-y(2x y + b))/(x^2-3) To match the desired form, we can factor out -1 from the denominator: (dy)/(dx) = (y(2x y + b))/(-(x^2-3)) (dy)/(dx) = (y(2x y + b))/(3-x^2) Problem 5i) b) Find the equation of the tangent to the curve at the point P(2, e^2). Step 1: Find the value of the constant b. Substitute the point (x, y) = (2, e^2) into the original curve equation (x^2-3) y + bx = 4: (2^2-3)(e^2) + b(2) = 4 (4-3)(2) + 2b = 4 1(2) + 2b = 4 2 + 2b = 4 2b = 2 b = 1 Step 2: Calculate the slope of the tangent at P(2, e^2). Use the derivative formula from part (a) with x=2, y=e^2, and b=1: m = (dy)/(dx) |_(2, e^2) = (e^2(2(2) (e^2) + 1))/(3-2^2) m = (e^2(4(2) + 1))/(3-4) m = (e^2(8 + 1))/(-1) m = (9e^2)/(-1) m = -9e^2 Step 3: Write the equation of the tangent line. Using the point-slope form Y - y_1 = m(X - x_1) with (x_1, y_1) = (2, e^2) and m = -9e^2: Y - e^2 = -9e^2(X - 2) Y - e^2 = -9e^2X + 18e^2 Y = -9e^2X + 18e^2 + e^2 Y = -9e^2X + 19e^2 Problem 5ii) Solve the differential equation (1+x^2)(dy)/(dx) = 2xy, given that y=1 when x=0. Step 1: Separate the variables. Divide both sides by y(1+x^2): (1)/(y)(dy)/(dx) = (2x)/(1+x^2) Multiply by dx: (1)/(y) dy = (2x)/(1+x^2) dx Step 2: Integrate both sides. (1)/(y) dy = (2x)/(1+x^2) dx The left side integrates to |y|. For the right side, let u = 1+x^2, so du = 2x\,dx. The integral becomes (1)/(u) du = |u| = (1+x^2) (since 1+x^2 is always positive). So, we have: |y| = (1+x^2) + C Step 3: Use the initial condition to find the constant C. Given y=1 when x=0: |1| = (1+0^2) + C 0 = (1) + C 0 = 0 + C C = 0 Step 4: Write the particular solution. Substitute C=0 back into the general solution: |y| = (1+x^2) Exponentiate both sides to solve for y: e^|y| = e^(1+x^2) |y| = 1+x^2 Since y=1 (positive) when x=0, and 1+x^2 is always positive, we can remove the absolute value: y = 1+x^2 What's next?