Let's solve each part step-by-step: --- ### **(a) Given:** \[ \sin 2x \frac{dy}{dx} = y \cos 2x \] with initial condition \( y\left(\frac{\pi}{4}\right) = 2 \). **Step 1:** Rearrange to separate variables: \[ \sin 2x \frac{dy}{dx} = y \cos 2x \] \[ \frac{dy}{dx} = \frac{y \cos 2x}{\sin 2x} \] \[ \frac{dy}{dx} = y \cot 2x \] \[ \frac{1}{y} dy = \cot 2x \, dx \] **Step 2:** Integrate both sides: \[ \int \frac{1}{y} dy = \int \cot 2x \, dx \] \[ \ln |y| = \int \frac{\cos 2x}{\sin 2x} dx \] Let \( u = \sin 2x \), \( du = 2 \cos 2x dx \implies \frac{1}{2} du = \cos 2x dx \). \[ \int \cot 2x \, dx = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln|u| = \frac{1}{2} \ln|\sin 2x| \] So, \[ \ln |y| = \frac{1}{2} \ln |\sin 2x| + C \] \[ |y| = e^C |\sin 2x|^{1/2} \] Let \( e^C = K \): \[ y = K (\sin 2x)^{1/2} \] **Step 3:** Apply the initial condition \( y\left(\frac{\pi}{4}\right) = 2 \): \[ y\left(\frac{\pi}{4}\right) = K (\sin (\frac{\pi}{2}))^{1/2} = K (1)^{1/2} = K \] So \( K = 2 \). **Final Answer:** \[ \boxed{y = 2 (\sin 2x)^{1/2}} \] --- ### **(b) Given:** \[ x \frac{dy}{dx} + 4y = 6x^2 - 5x \] with \( y(1) = 1 \). **Step 1:** Rewrite in standard linear form: \[ \frac{dy}{dx} + \frac{4}{x} y = 6x - 5 \] **Step 2:** Find integrating factor (IF): \[ IF = e^{\int \frac{4}{x} dx} = e^{4 \ln x} = x^4 \] **Step 3:** Multiply both sides by IF: \[ x^4 \frac{dy}{dx} + 4x^3 y = x^4 (6x - 5) \] \[ x^4 \frac{dy}{dx} + 4x^3 y = 6x^5 - 5x^4 \] Left side is derivative: \[ \frac{d}{dx}(x^4 y) = 6x^5 - 5x^4 \] **Step 4:** Integrate both sides: \[ \int \frac{d}{dx}(x^4 y) dx = \int (6x^5 - 5x^4) dx \] \[ x^4 y = x^6 - x^5 + C \] **Step 5:** Solve for \( y \): \[ y = x^2 - x + \frac{C}{x^4} \] **Step 6:** Apply initial condition \( y(1) = 1 \): \[ 1 = 1 - 1 + C \implies C = 1 \] **Final Answer:** \[ \boxed{y = x^2 - x + \frac{1}{x^4}} \] --- ### **(c) Given:** \[ (2y - x) \frac{dy}{dx} = y \] **(i) Show exactness** Write as: \[ (2y - x)\frac{dy}{dx} - y = 0 \] \[ (2y - x)\frac{dy}{dx} - y = 0 \] Let \( M(x, y) = -y \), \( N(x, y) = 2y - x \) Check exactness: \[ \frac{\partial M}{\partial x} = 0,\quad \frac{\partial N}{\partial y} = 2 \] Not exact, but let's check the form. Instead, solve as separable. **(ii) Find general solution** Rewrite: \[ (2y - x) \frac{dy}{dx} = y \] \[ (2y - x) dy = y dx \] \[ 2y dy - x dy = y dx \] \[ 2y dy - y dx = x dy \] \[ 2y dy - y dx - x dy = 0 \] \[ - y dx + (2y - x) dy = 0 \] Now, integrate both sides. Rearranged as: \[ (2y - x) dy = y dx \] \[ \frac{2y - x}{y} dy = dx \] \[ 2 dy - \frac{x}{y} dy = dx \] But let's separate variables: \[ (2y - x) \frac{dy}{dx} = y \] \[ (2y - x) \frac{dy}{dx} - y = 0 \] Let’s try integrating factor method, but best to separate: \[ (2y - x) \frac{dy}{dx} = y \] \[ (2y - x) dy = y dx \] \[ 2y dy - x dy = y dx \] \[ 2y dy - y dx = x dy \] \[ 2y dy - y dx - x dy = 0 \] \[ - y dx + (2y - x) dy = 0 \] Let’s write as: \[ - y dx + (2y - x) dy = 0 \] This is exact if \( \frac{\partial}{\partial y}(-y) = -1 \), \( \frac{\partial}{\partial x}(2y - x) = -1 \), so it is exact. So, yes, it is an exact ODE. **(iii) Solution** Find \( F(x, y) \): \[ \frac{\partial F}{\partial x} = -y \implies F(x, y) = -xy + g(y) \] \[ \frac{\partial F}{\partial y} = -x + g'(y) = 2y - x \implies g'(y) = 2y \implies g(y) = y^2 \] Thus, \[ F(x, y) = -xy + y^2 = C \] **(iv) Apply initial condition \( y(2) = 3 \):** \[ -(2)(3) + 3^2 = C \] \[ -6 + 9 = C \] \[ C = 3 \] --- ### **Final Answers:** #### (a) \[ \boxed{y = 2 (\sin 2x)^{1/2}} \] #### (b) \[ \boxed{y = x^2 - x + \frac{1}{x^4}} \] #### (c) (i) The ODE is exact. (ii) General solution: \(\boxed{-xy + y^2 = C}\) (iii) \(\boxed{C = 3}\)