Let's solve each part step-by-step: --- ## (a) **Exactness and Solution** Given: \[ (2x e^y) dx + (x^2 e^y + \cos y) dy = 0 \] Let \( M(x, y) = 2x e^y \) and \( N(x, y) = x^2 e^y + \cos y \). ### **Step 1: Check for Exactness** A differential equation \( M(x, y) dx + N(x, y) dy = 0 \) is **exact** if: \[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \] Calculate: - \( \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(2x e^y) = 2x e^y \) - \( \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(x^2 e^y + \cos y) = 2x e^y \) Since \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), **the equation is exact.** --- ### **Step 2: Find the Solution** There exists a function \( F(x, y) \) such that: \[ \frac{\partial F}{\partial x} = M(x, y) \quad \text{and} \quad \frac{\partial F}{\partial y} = N(x, y) \] Integrate \( M(x, y) \) with respect to \( x \): \[ F(x, y) = \int 2x e^y dx = x^2 e^y + h(y) \] Now, differentiate with respect to \( y \): \[ \frac{\partial F}{\partial y} = x^2 e^y + h'(y) \] Set equal to \( N(x, y) \): \[ x^2 e^y + h'(y) = x^2 e^y + \cos y \implies h'(y) = \cos y \implies h(y) = \sin y \] So, \[ F(x, y) = x^2 e^y + \sin y \] **General solution:** \[ \boxed{x^2 e^y + \sin y = C} \] --- ## (b) **Find \( k \) for Exactness** ### **(i)** Given: \[ M = y^3 + k x y^4 - 2x, \quad N = 3x y^2 + 20x^2 y^3 \] Calculate: - \( \frac{\partial M}{\partial y} = 3y^2 + 4k x y^3 \) - \( \frac{\partial N}{\partial x} = 3y^2 + 40x y^3 \) Set equal for exactness: \[ 3y^2 + 4k x y^3 = 3y^2 + 40x y^3 \implies 4k x y^3 = 40x y^3 \implies 4k = 40 \implies k = 10 \] **Final answer:** \(\boxed{k = 10}\) --- ### **(ii)** Given: \[ M = 6x y^3 + \cos y, \quad N = 2k x^2 y^2 - x \sin y \] Calculate: - \( \frac{\partial M}{\partial y} = 18x y^2 - \sin y \) - \( \frac{\partial N}{\partial x} = 4k x y^2 - \sin y \) Set equal for exactness: \[ 18x y^2 - \sin y = 4k x y^2 - \sin y \implies 18x y^2 = 4k x y^2 \implies 18 = 4k \implies k = \frac{9}{2} \] **Final answer:** \(\boxed{k = \dfrac{9}{2}}\) --- ## (c) **Initial Value Problem** \[ \frac{dy}{dx} = \frac{xy^2 - \cos x \sin x}{y(1-x^2)}, \qquad y(0) = 2 \] ### **Step 1: Rearrangement** \[ \frac{dy}{dx} = \frac{xy^2}{y(1-x^2)} - \frac{\cos x \sin x}{y(1-x^2)} = \frac{x y}{1-x^2} - \frac{\cos x \sin x}{y(1-x^2)} \] Bring \( y \) terms to one side: \[ y \, \frac{dy}{dx} = \frac{x y^2}{1-x^2} - \frac{\cos x \sin x}{1-x^2} \] Write as: \[ y \frac{dy}{dx} - \frac{x y^2}{1-x^2} = -\frac{\cos x \sin x}{1-x^2} \] Alternatively, bring all terms involving \( y \) to one side: \[ \frac{dy}{dx} - \frac{x}{1-x^2} y = -\frac{\cos x \sin x}{y(1-x^2)} \] Multiply both sides by \( y \): \[ y \frac{dy}{dx} - \frac{x}{1-x^2} y^2 = -\frac{\cos x \sin x}{1-x^2} \] Let’s separate variables if possible: \[ y \frac{dy}{dx} = \frac{x y^2 - \cos x \sin x}{1-x^2} \] Now, write as: \[ y \frac{dy}{dx} = \frac{x y^2}{1-x^2} - \frac{\cos x \sin x}{1-x^2} \] Let’s try to separate variables: \[ y \frac{dy}{dx} - \frac{x y^2}{1-x^2} = - \frac{\cos x \sin x}{1-x^2} \] Now, \( y \frac{dy}{dx} - \frac{x}{1-x^2} y^2 \) is a linear ODE in \( y^2 \). Let’s use substitution: Let \( u = y^2 \implies \frac{du}{dx} = 2y \frac{dy}{dx} \implies y \frac{dy}{dx} = \frac{1}{2} \frac{du}{dx} \) Substitute: \[ \frac{1}{2} \frac{du}{dx} - \frac{x}{1-x^2} u = -\frac{\cos x \sin x}{1-x^2} \] Multiply both sides by 2: \[ \frac{du}{dx} - \frac{2x}{1-x^2} u = -\frac{2 \cos x \sin x}{1-x^2} \] This is a linear ODE in \( u \): \[ \frac{du}{dx} + P(x) u = Q(x) \] where \( P(x) = -\frac{2x}{1-x^2} \), \( Q(x) = -\frac{2 \cos x \sin x}{1-x^2} \) ### **Step 2: Find Integrating Factor** \[ \text{IF} = \exp\left(\int -\frac{2x}{1-x^2} dx \right) \] Let \( 1-x^2 = t \implies dt = -2x dx \implies -\frac{1}{2} dt = x dx \) So, \[ \int -\frac{2x}{1-x^2} dx = \int \frac{2x}{x^2-1} dx \] But let's be careful with the sign: \[ -\frac{2x}{1-x^2} = \frac{2x}{x^2-1} \] So, \[ \int \frac{2x}{x^2-1} dx \] Let \( s = x^2-1 \implies ds = 2x dx \implies \frac{ds}{s} = dx \) Therefore, \[ \int \frac{2x}{x^2-1} dx = \int \frac{ds}{s} = \ln|x^2-1| \] So, \[ \text{Integrating factor} = e^{\ln|x^2-1|} = |x^2-1| \] Assume \( x^2<1 \) for initial value \( x=0 \), so \( x^2-1 < 0 \), but integrating factor can be \( x^2-1 \). ### **Step 3: Multiply Through by Integrating Factor** Multiply both sides by \( x^2-1 \): \[ (x^2-1) \frac{du}{dx} - 2x u = -2 \cos x \sin x \] But recall, integrating factor method: Multiply both sides by \( x^2-1 \): \[ \frac{d}{dx}[(x^2-1) u] = -2 \cos x \sin x \] ### **Step 4: Integrate Both Sides** \[ \frac{d}{dx} [(x^2-1) u] = -2 \cos x \sin x \] \[ (x^2-1) u = \int -2 \cos x \sin x dx + C \] Recall: \[ -2 \cos x \sin x = -\sin 2x \] So, \[ \int -2 \cos x \sin x dx = \int -\sin 2x dx = \frac{\cos 2x}{2} \] Thus, \[ (x^2-1) u = \frac{\cos 2x}{2} + C \] Recall \( u = y^2 \): \[ (x^2-1) y^2 = \frac{\cos 2x}{2} + C \] ### **Step 5: Apply Initial Condition \( y(0) = 2 \)** At \( x = 0 \), \( y = 2 \): \[ (0^2-1)(2^2) = \frac{\cos 0}{2} + C \] \[ (-1)(4) = \frac{1}{2} + C \] \[ -4 = \frac{1}{2} + C \] \[ C = -4 - \frac{1}{2} = -\frac{9}{2} \] ### **Step 6: Write Final Solution** \[ (x^2-1) y^2 = \frac{\cos 2x}{2} - \frac{9}{2} \] Or, \[ \boxed{ (x^2-1) y^2 = \frac{\cos 2x - 9}{2} } \] Or, solving for \( y \): \[ \boxed{ y(x) = \pm \sqrt{\frac{\cos 2x - 9}{2(x^2-1)}} } \] But considering initial value \( y(0) = 2 > 0 \), take the positive root. --- ## **Summary of Final Answers** ### (a) \[ \boxed{x^2 e^y + \sin y = C} \] ### (b) (i) \( \boxed{k = 10} \) (ii) \( \boxed{k = \frac{9}{2}} \) ### (c) \[ \boxed{ y(x) = \sqrt{\frac{\cos 2x - 9}{2(x^2-1)}} } \] with initial condition \( y(0) = 2 \).