1. **Problem 14:** Given the differential equation $ (2xy + y^2) dx + (2x^2 + 3xy) dy = 0 $ and an integrating factor of the form $ \mu(x,y) = x^m y^n $, find $m$ and $n$ so the equation becomes exact, then solve it. 2. **Step 1:** Check if the original equation is exact. Calculate $M = 2xy + y^2$ and $N = 2x^2 + 3xy$. Compute partial derivatives: $$ \frac{\partial M}{\partial y} = 2x + 2y, \quad \frac{\partial N}{\partial x} = 4x + 3y $$ Since $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$, the equation is not exact. 3. **Step 2:** Find integrating factor $\mu = x^m y^n$. The equation becomes exact if: $$ \frac{\partial}{\partial y}(\mu M) = \frac{\partial}{\partial x}(\mu N) $$ Calculate: $$ \frac{\partial}{\partial y}(x^m y^n (2xy + y^2)) = \frac{\partial}{\partial y}(2x^{m+1} y^{n+1} + x^m y^{n+2}) $$ $$ = 2x^{m+1} (n+1) y^n + x^m (n+2) y^{n+1} $$ Similarly, $$ \frac{\partial}{\partial x}(x^m y^n (2x^2 + 3xy)) = \frac{\partial}{\partial x}(2x^{m+2} y^n + 3x^{m+1} y^{n+1}) $$ $$ = 2(m+2) x^{m+1} y^n + 3(m+1) x^m y^{n+1} $$ Equate coefficients of $x^{m+1} y^n$ and $x^m y^{n+1}$: $$ 2(n+1) = 2(m+2) \implies n+1 = m+2 \implies n = m+1 $$ $$ n+2 = 3(m+1) $$ Substitute $n = m+1$ into second: $$ m+1 + 2 = 3(m+1) \implies m+3 = 3m + 3 \implies 0 = 2m \implies m=0 $$ Then $n = m+1 = 1$. 4. **Step 3:** Integrating factor is $\mu = x^0 y^1 = y$. Multiply original equation by $y$: $$ y(2xy + y^2) dx + y(2x^2 + 3xy) dy = (2xy^2 + y^3) dx + (2x^2 y + 3x y^2) dy = 0 $$ Check exactness: $$ M = 2xy^2 + y^3, \quad N = 2x^2 y + 3x y^2 $$ $$ \frac{\partial M}{\partial y} = 4xy + 3y^2, \quad \frac{\partial N}{\partial x} = 4xy + 3y^2 $$ Exact! 5. **Step 4:** Find potential function $\Psi(x,y)$ such that: $$ \frac{\partial \Psi}{\partial x} = M = 2xy^2 + y^3 $$ Integrate w.r.t. $x$: $$ \Psi = \int (2xy^2 + y^3) dx = x^2 y^2 + x y^3 + h(y) $$ Differentiate w.r.t. $y$: $$ \frac{\partial \Psi}{\partial y} = 2x^2 y + 3x y^2 + h'(y) $$ Set equal to $N$: $$ 2x^2 y + 3x y^2 + h'(y) = 2x^2 y + 3x y^2 \implies h'(y) = 0 $$ So $h(y)$ is constant. 6. **Step 5:** General solution: $$ \Psi(x,y) = C \implies x^2 y^2 + x y^3 = C $$ --- 7. **Problem 15:** Given $$ (2 \sin^3 y + 3x \sin y) dx + (4x \cos y \sin^2 y + 2x^2 \cos y) dy = 0 $$ with integrating factor $\mu(x,y) = x f(y)$, find $f(y)$. 8. **Step 1:** Let $\mu = x f(y)$. Multiply original equation: $$ M = (2 \sin^3 y + 3x \sin y) x f(y) = 2x f(y) \sin^3 y + 3x^2 f(y) \sin y $$ $$ N = (4x \cos y \sin^2 y + 2x^2 \cos y) x f(y) = 4x^2 f(y) \cos y \sin^2 y + 2x^3 f(y) \cos y $$ 9. **Step 2:** For exactness: $$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$ Calculate: $$ \frac{\partial M}{\partial y} = 2x f'(y) \sin^3 y + 6x f(y) \sin^2 y \cos y + 3x^2 f'(y) \sin y + 3x^2 f(y) \cos y $$ $$ \frac{\partial N}{\partial x} = 8x f(y) \cos y \sin^2 y + 6x^2 f(y) \cos y $$ Equate coefficients of $x$ and $x^2$ terms: For $x$ terms: $$ 2 f'(y) \sin^3 y + 6 f(y) \sin^2 y \cos y = 8 f(y) \cos y \sin^2 y $$ Simplify: $$ 2 f'(y) \sin^3 y = 2 f(y) \cos y \sin^2 y $$ $$ f'(y) \sin y = f(y) \cos y $$ 10. **Step 3:** Solve ODE for $f(y)$: $$ \frac{f'(y)}{f(y)} = \frac{\cos y}{\sin y} $$ Integrate: $$ \ln |f(y)| = \ln |\sin y| + C \implies f(y) = K \sin y $$ Ignore constant $K$ (absorbed in integrating factor), so: $$ f(y) = \sin y $$ 11. **Step 4:** Integrating factor: $$ \mu = x \sin y $$ Multiply original equation by $x \sin y$ and solve similarly (not requested here). --- 12. **Problem 16:** Cooling problem. Given temperature $\theta$ of a body cooling in a room at $18^\circ C$, with initial $\theta_0 = 70$, after 5 minutes $\theta = 57$. Find time to cool to $40$. 13. **Step 1:** Newton's law of cooling: $$ \frac{d\theta}{dt} = -k (\theta - 18) $$ 14. **Step 2:** Solve ODE: $$ \frac{d\theta}{\theta - 18} = -k dt $$ Integrate: $$ \ln|\theta - 18| = -k t + C $$ At $t=0$, $\theta=70$: $$ \ln(70 - 18) = C = \ln 52 $$ So: $$ \ln|\theta - 18| = -k t + \ln 52 $$ $$ |\theta - 18| = 52 e^{-k t} $$ 15. **Step 3:** Use $t=5$, $\theta=57$: $$ 57 - 18 = 39 = 52 e^{-5k} $$ $$ e^{-5k} = \frac{39}{52} = \frac{3}{4} $$ Take ln: $$ -5k = \ln \frac{3}{4} \implies k = -\frac{1}{5} \ln \frac{3}{4} $$ 16. **Step 4:** Find time $t$ for $\theta=40$: $$ 40 - 18 = 22 = 52 e^{-k t} $$ $$ e^{-k t} = \frac{22}{52} = \frac{11}{26} $$ Take ln: $$ -k t = \ln \frac{11}{26} $$ Substitute $k$: $$ t = -\frac{\ln \frac{11}{26}}{k} = -\frac{\ln \frac{11}{26}}{-\frac{1}{5} \ln \frac{3}{4}} = 5 \frac{\ln \frac{11}{26}}{\ln \frac{3}{4}} $$ Calculate numerically: $$ \ln \frac{11}{26} \approx \ln 0.4231 = -0.8602 $$ $$ \ln \frac{3}{4} \approx \ln 0.75 = -0.2877 $$ So: $$ t \approx 5 \times \frac{-0.8602}{-0.2877} = 5 \times 2.99 = 14.95 \text{ minutes} $$ 17. **Step 5:** Time elapsed is 5 minutes already, so additional time needed: $$ 14.95 - 5 = 9.95 \text{ minutes} $$ --- **Final answers:** 14. $m=0$, $n=1$, solution: $x^2 y^2 + x y^3 = C$ 15. $f(y) = \sin y$, integrating factor $\mu = x \sin y$ 16. Additional cooling time to reach $40^\circ C$ is approximately 9.95 minutes.