Problem 11: Verify Exactness and Find the General Solution

Given the differential equation: $(1 + y e^{xy}) \, dx + (2y + x e^{xy}) \, dy = 0$

Step 1: Verify Exactness

To verify if the given equation is exact, we need to check if: $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$

Here, $M(x,y) = 1 + y e^{xy}$ and $N(x,y) = 2y + x e^{xy}$.

Calculate $\frac{\partial M}{\partial y}$: $\frac{\partial M}{\partial y} = \frac{\partial}{\partial y} (1 + y e^{xy}) = e^{xy} + xy e^{xy}$

Calculate $\frac{\partial N}{\partial x}$: $\frac{\partial N}{\partial x} = \frac{\partial}{\partial x} (2y + x e^{xy}) = y e^{xy} + x y e^{xy} = e^{xy}(y + xy)$

Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} = e^{xy}(1 + xy)$, the equation is exact.

Step 2: Find the General Solution

The general solution can be found by integrating $M(x,y)$ with respect to $x$ and $N(x,y)$ with respect to $y$:

$\int M(x,y) \, dx = \int (1 + y e^{xy}) \, dx = x + \frac{e^{xy}}{x} + h(y)$ $\int N(x,y) \, dy = \int (2y + x e^{xy}) \, dy = y^2 + \frac{e^{xy}}{y} + g(x)$

Combining these, we get the potential function $\psi(x,y)$: $\psi(x,y) = x + y e^{xy} + C$

Thus, the general solution is: $x + y e^{xy} = C$

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Problem 12: Solve the Nonhomogeneous System in Vector Form

Given the nonhomogeneous system of linear equations with the given augmented matrix RREF:

1 & 0 & -2 & 3 & 2 & | & -24 \\ 0 & 1 & -2 & 0 & 1 & | & -7 \\ 0 & 0 & 0 & 1 & 0 & | & 4 \\ \end{bmatrix}$$ Express the solution in vector form: Let $$x_3 = t$$ and $$x_5 = s$$, where $$t$$ and $$s$$ are free parameters. From the RREF: $$x_1 - 2x_3 + 3x_4 + 2x_5 = -24$$ $$x_2 - 2x_3 + x_5 = -7$$ $$x_4 = 4$$ Thus: $$x_1 = 2t - 3(4) - 2s - 24 = 2t - 2s - 36$$ $$x_2 = 2t - s - 7$$ $$x_4 = 4$$ $$x_3 = t$$ $$x_5 = s$$ So, the solution in vector form is: $$\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix} = \begin{bmatrix} 2t - 2s - 36 \\ 2t - s - 7 \\ t \\ 4 \\ s \end{bmatrix} = t \begin{bmatrix} 2 \\ 2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + s \begin{bmatrix} -2 \\ -1 \\ 0 \\ 0 \\ 1 \end{bmatrix} + \begin{bmatrix} -36 \\ -7 \\ 0 \\ 4 \\ 0 \end{bmatrix}$$ --- ### Problem 13: Find the General Solution Given: $$xy' + 6y = 3x y^{4/3}$$ This is a Bernoulli equation. Rewrite it as: $$y' + \frac{6}{x} y = 3 y^{4/3}$$ Let $$v = y^{1/3}$$. Then $$y = v^3$$ and $$y' = 3v^2 v'$$. Substituting these into the equation: $$3v^2 v' + \frac{6}{x} v^3 = 3 v^4$$ \[ v' + \frac{2