1. **Problem 1: Solve the initial value problem** $$ (e^{x+y} + y e^{y}) \, dx + (x e^{y} - 1) \, dy = 0, \quad y(0) = -1 $$ Step 1: Identify if the differential equation is exact. Let $$M = e^{x+y} + y e^{y}, \quad N = x e^{y} - 1.$$ Check if $$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}.$$ Calculate $$\frac{\partial M}{\partial y} = e^{x+y} + e^{y} + y e^{y},$$ Calculate $$\frac{\partial N}{\partial x} = e^{y}.$$ Since $$\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x},$$ the equation is not exact. Step 2: Try an integrating factor depending on $$x$$ or $$y$$. Check $$\frac{\partial}{\partial y}\left(\frac{M}{N}\right)$$ or $$\frac{\partial}{\partial x}\left(\frac{N}{M}\right)$$ to find an integrating factor. Step 3: We find an integrating factor $$\mu = e^{-y}$$ since multiplying entire equation by $$e^{-y}$$ gives: $$ \tilde{M} = e^{x} + y, \quad \tilde{N} = x - e^{-y}.$$ Check exactness again: $$\frac{\partial \tilde{M}}{\partial y} = 1, \quad \frac{\partial \tilde{N}}{\partial x} = 1,$$ so the new equation is exact. Step 4: Find the potential function $$\Psi(x,y)$$ such that $$\frac{\partial \Psi}{\partial x} = e^{x} + y,$$ Integrate w.r.t. $$x$$: $$\Psi(x,y) = e^{x} + x y + h(y).$$ Step 5: Use $$\frac{\partial \Psi}{\partial y} = x - e^{-y}$$: $$\frac{\partial \Psi}{\partial y} = x + h'(y) = x - e^{-y} \implies h'(y) = - e^{-y}.$$ Integrate: $$h(y) = e^{-y} + C.$$ Step 6: The implicit solution is $$\Psi(x,y) = e^{x} + xy + e^{-y} = C.$$ Step 7: Use initial condition $$y(0) = -1$$: $$e^{0} + 0 \cdot (-1) + e^{1} = C \implies 1 + e = C.$$ **Final solution:** $$e^{x} + xy + e^{-y} = 1 + e.$$ --- 2. **Problem 2: Find the particular solution of** $$y y'' - 2y y' \ln y = (y')^{2}, \quad y(0) = 1, \quad y'(0) = 1.$$ Step 1: Rewrite the equation: $$y y'' - 2 y y' \ln y - (y')^{2} = 0.$$ Step 2: Use substitution $$p = y', \quad p' = y'',$$ so the equation becomes $$y p' - 2 y p \ln y - p^{2} = 0.$$ Step 3: Divide by $$y$$ (assuming $$y \neq 0$$): $$p' - 2 p \ln y - \frac{p^{2}}{y} = 0.$$ Step 4: Try substitution to reduce the order or solve using known methods. Consider the form suggests substitution about \(p\) and \(y\). Rewrite the original as: $$y y'' = 2 y y' \ln y + (y')^{2}.$$ Step 5: Observe the structure, try change of variable: Define $$u = \ln y$$; then $$y = e^{u}$$. Then, $$y' = y u' = e^{u} u',$$ $$y'' = \frac{d}{dx}(y') = \frac{d}{dx} (e^{u} u') = e^{u} u' u' + e^{u} u'' = e^{u} (u'^{2} + u'').$$ Step 6: Substitute into the equation: Left side: $$ y y'' = e^{u} \times e^{u} (u'' + u'^{2}) = e^{2u} (u'' + u'^{2}).$$ Right side: $$ 2 y y' \ln y + (y')^{2} = 2 e^{u} \times e^{u} u' \times u + (e^{u} u')^{2} = 2 e^{2u} u u' + e^{2u} u'^{2}.$$ So equation becomes: $$ e^{2u} (u'' + u'^{2}) = 2 e^{2u} u u' + e^{2u} u'^{2}.$$ Divide both sides by $$e^{2u}$$: $$ u'' + u'^{2} = 2 u u' + u'^{2}.$$ Simplify: $$ u'' = 2 u u' .$$ Step 7: This is a second order nonlinear ODE for $$u$$: $$ u'' = 2 u u' .$$ Step 8: Let $$v = u'$$, so $$u'' = v' = \frac{dv}{dx}$$. Use chain rule: $$v' = \frac{dv}{du} \frac{du}{dx} = v \frac{dv}{du}.$$ Equation becomes $$ v \frac{dv}{du} = 2 u v \, \Rightarrow \, \frac{dv}{du} = 2 u.$$ Step 9: Integrate w.r.t. $$u$$: $$ v = \int 2 u \, du = u^{2} + C_{1}.$$ Step 10: Recall $$v = u' = \frac{du}{dx}$$, so $$ \frac{du}{dx} = u^{2} + C_{1}.$$ Step 11: Use initial conditions: when $$x=0$$, $$y=1$$, so $$u(0) = \ln 1=0$$. Also, $$y'(0) =1$$, so $$y' = e^{u} u'$$, $$1 = y'(0) = e^{u(0)} u'(0) = e^{0} u'(0) = u'(0) = v(0).$$ From step 9, $$v(0) = u^{2}(0) + C_{1} = 0 + C_{1} = C_{1}$$, so $$C_{1} = 1$$. Step 12: We have $$\frac{du}{dx} = u^{2} + 1.$$ Separate variables: $$ \frac{du}{u^{2} + 1} = dx.$$ Step 13: Integrate: $$ \int \frac{du}{u^{2} + 1} = \int dx,$$ which gives $$ \arctan(u) = x + C_{2}.$$ Step 14: Use initial condition $$u(0) = 0$$: $$ \arctan(0) = 0 = 0 + C_{2} \, \Rightarrow \, C_{2} = 0.$$ Step 15: Solve for $$u$$: $$ u = \tan x.$$ Step 16: Recall $$y = e^{u} = e^{\tan x}.$$ **Final solution:** $$ y = e^{\tan x}.$$