Let's go through each problem one by one with step-by-step solutions and explanations. --- ## **2(a).** \[ \frac{dy}{dx} = e^{-y} + e^{2x-y} \] **Step 1: Factor \( e^{-y} \):** \[ e^{-y} + e^{2x-y} = e^{-y}(1 + e^{2x}) \] **Step 2: Separate variables:** \[ \frac{dy}{dx} = e^{-y}(1 + e^{2x}) \] \[ e^{y}\,dy = (1 + e^{2x})\,dx \] **Step 3: Integrate both sides:** \[ \int e^{y}\,dy = \int (1 + e^{2x})\,dx \] \[ e^{y} = x + \frac{1}{2}e^{2x} + C \] **Final Answer:** \[ \boxed{e^{y} = x + \frac{1}{2}e^{2x} + C} \] --- ## **3(a).** \[ \begin{cases} (\sin(x + y^2) + e^{3x} - 7)\,dx + (2y\sin(x + y^2) + y - 1)\,dy = 0 \\ y(0) = 0 \end{cases} \] **Step 1: Check if the equation is exact.** Let \( M(x,y) = \sin(x+y^2) + e^{3x} - 7 \), \( N(x,y) = 2y\sin(x+y^2) + y - 1 \). \[ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}\left(\sin(x+y^2)\right) = 2y\cos(x+y^2) \] \[ \frac{\partial N}{\partial x} = 2y\cos(x+y^2) \] They are equal, so the equation is **exact**. **Step 2: Find potential function \( F(x,y) \):** \[ F_x = M, \quad F_y = N \] Integrate \( M \) with respect to \( x \): \[ F(x, y) = \int (\sin(x+y^2) + e^{3x} - 7) dx \] \[ = -\cos(x+y^2) + \frac{1}{3}e^{3x} - 7x + h(y) \] Differentiate with respect to \( y \) and set equal to \( N \): \[ F_y = 2y\sin(x+y^2) + h'(y) \] Set equal to \( N \): \[ 2y\sin(x+y^2) + h'(y) = 2y\sin(x+y^2) + y - 1 \implies h'(y) = y - 1 \implies h(y) = \frac{1}{2}y^2 - y \] **So,** \[ F(x, y) = -\cos(x+y^2) + \frac{1}{3}e^{3x} - 7x + \frac{1}{2}y^2 - y \] The solution is: \[ F(x, y) = C \] Substitute \( y(0) = 0 \): \[ -\cos(0 + 0^2) + \frac{1}{3}e^{0} - 7(0) + \frac{1}{2}(0)^2 - 0 = C \implies -1 + \frac{1}{3} = C \implies C = -\frac{2}{3} \] **Final Answer:** \[ \boxed{-\cos(x+y^2) + \frac{1}{3}e^{3x} - 7x + \frac{1}{2}y^2 - y = -\frac{2}{3}} \] --- ## **3(b).** \[ \frac{dy}{dx} = \frac{x + 2y}{2x + y} \] **Step 1: Substitute \( u = x + 2y \).** Let’s try \( y = vx \) (homogeneous equation): Rewrite: \[ \frac{dy}{dx} = \frac{x + 2y}{2x + y} \] Let \( y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx} \) \[ v + x\frac{dv}{dx} = \frac{x + 2vx}{2x + vx} = \frac{x(1 + 2v)}{x(2 + v)} = \frac{1 + 2v}{2 + v} \] \[ x\frac{dv}{dx} = \frac{1 + 2v}{2 + v} - v = \frac{1 + 2v - v(2 + v)}{2 + v} = \frac{1 + 2v - 2v - v^2}{2 + v} = \frac{1 - v^2}{2 + v} \] So: \[ \frac{2 + v}{1 - v^2} dv = \frac{dx}{x} \] **Step 2: Separate and integrate** \[ \frac{2 + v}{1 - v^2} = \frac{2}{1 - v^2} + \frac{v}{1 - v^2} \] \[ \frac{2}{1 - v^2} = \frac{1}{1 - v} + \frac{1}{1 + v} \] \[ \frac{v}{1 - v^2} = \frac{1}{2} \left( \frac{1}{1 - v} - \frac{1}{1 + v} \right) \] So, \[ \frac{2 + v}{1 - v^2} = \frac{1}{1 - v} + \frac{1}{1 + v} + \frac{1}{2}\left(\frac{1}{1 - v} - \frac{1}{1 + v}\right) = \frac{1 + \frac{1}{2}}{1 - v} + \frac{1 - \frac{1}{2}}{1 + v} = \frac{3/2}{1 - v} + \frac{1/2}{1 + v} \] Integrate: \[ \int \left( \frac{3/2}{1 - v} + \frac{1/2}{1 + v} \right) dv = \int \frac{dx}{x} \] \[ -\frac{3}{2} \ln|1 - v| + \frac{1}{2} \ln|1 + v| = \ln|x| + C \] **Step 3: Substitute back \( v = y/x \):** \[ -\frac{3}{2} \ln|1 - y/x| + \frac{1}{2} \ln|1 + y/x| = \ln|x| + C \] Or rearrange as: \[ \ln \left| \frac{(1 + y/x)^{1/2}}{(1 - y/x)^{3/2}} \right| = \ln|x| + C \] Exponentiate both sides: \[ \frac{(1 + y/x)^{1/2}}{(1 - y/x)^{3/2}} = K x \] or equivalently: \[ \boxed{\frac{\sqrt{1 + y/x}}{(1 - y/x)^{3/2}} = Cx} \] --- ## **4(a).** \[ (x-1)^2 \frac{dy}{dx} + (x-1)y = x+1 \] **Step 1: Divide both sides by \((x-1)^2\):** \[ \frac{dy}{dx} + \frac{1}{x-1}y = \frac{x+1}{(x-1)^2} \] This is a linear ODE. **Step 2: Find integrating factor:** \[ \mu(x) = \exp\left(\int \frac{1}{x-1}dx\right) = |x-1| \] (For \(x > 1\), \(x-1\). For \(x < 1\), \(-(x-1)\). Consider \(x-1\) for now.) Multiply both sides by \(x-1\): \[ (x-1)\frac{dy}{dx} + y = \frac{x+1}{x-1} \] **Step 3: Recognize left side as derivative:** \[ \frac{d}{dx} \left( y(x-1) \right ) = \frac{x+1}{x-1} \] **Step 4: Integrate both sides:** \[ y(x-1) = \int \frac{x+1}{x-1} dx \] \[ \frac{x+1}{x-1} = 1 + \frac{2}{x-1} \] So: \[ y(x-1) = \int \left( 1 + \frac{2}{x-1} \right) dx = x + 2\ln|x-1| + C \] \[ y = \frac{x + 2\ln|x-1| + C}{x-1} \] **Final Answer:** \[ \boxed{y(x) = \frac{x + 2\ln|x-1| + C}{x-1}} \] --- **4(b). Largest interval:** The solution and coefficients are not defined at \(x=1\). So the largest interval is \(\boxed{(-\infty, 1)}\) or \(\boxed{(1, \infty)}\) depending on initial conditions. --- **4(c). Transient term:** The transient term is the term that vanishes as \(x \to \infty\), i.e., the term containing \(C\). \[ \boxed{\text{Transient term: } \frac{C}{x-1}} \] --- If you need solutions for other parts, let me know!

Certainly! Here are the final answers for all parts: --- **2(a).** **General solution:** \[ \boxed{e^{y} = x + \frac{1}{2}e^{2x} + C} \] **Largest interval where solution exists:** \[ \boxed{(-\infty, \infty)} \] **Transient term (if any):** \[ \text{No transient term} \quad (\text{since the solution is explicit}) \] --- **3(a).** **Solution of the IVP:** \[ \boxed{-\cos(x + y^2) + \frac{1}{3}e^{3x} - 7x + \frac{1}{2} y^2 - y = -\frac{2}{3} \] **Initial condition \( y(0)=0 \) verified:** Yes. --- **3(b).** **General solution:** \[ \boxed{\frac{\sqrt{1 + y/x}}{(1 - y/x)^{3/2}} = C x} \] --- **4(a).** **Solution:** \[ \boxed{ y(x) = \frac{x + 2 \ln|x-1| + C}{x-1} } \] **Largest interval of existence:** \[ \boxed{(-\infty, 1)} \quad \text{or} \quad (1, \infty), \text{depending on initial conditions} \] **Transient term:** \[ \boxed{\frac{C}{x-1}} \] --- Let me know if you'd like any further clarifications!