Mathematics & Numerical Analysis

Separating the variables:

Integrating both sides:

Apply the initial condition $y(0) = 7$:

Given $\theta = t^n e^{-r^2/(4t)}$. Let $u = -r^2/(4t)$, so $\theta = t^n e^u$.
Calculate spatial derivatives:
$\frac{\partial \theta}{\partial r} = t^n e^u \left(-\frac{2r}{4t}\right) = -\frac{r}{2} t^{n-1} e^u$.
Then $r^2 \frac{\partial \theta}{\partial r} = -\frac{1}{2} t^{n-1} r^3 e^u$.
Differentiating with respect to $r$:
$\frac{\partial}{\partial r}\left(r^2 \frac{\partial \theta}{\partial r}\right) = -\frac{1}{2} t^{n-1} \left[3r^2 e^u + r^3 e^u \left(-\frac{r}{2t}\right)\right] = -\frac{1}{2} t^{n-1} e^u \left[3r^2 - \frac{r^4}{2t}\right]$.
Dividing by $r^2$:
$\text{LHS} = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial \theta}{\partial r}\right) = e^u \left[-\frac{3}{2} t^{n-1} + \frac{r^2}{4} t^{n-2}\right]$.
Now calculate time derivative:
$\text{RHS} = \frac{\partial \theta}{\partial t} = n t^{n-1} e^u + t^n e^u \left(\frac{r^2}{4t^2}\right) = e^u \left[n t^{n-1} + \frac{r^2}{4} t^{n-2}\right]$.
Equating LHS and RHS:
$-\frac{3}{2} t^{n-1} + \frac{r^2}{4} t^{n-2} = n t^{n-1} + \frac{r^2}{4} t^{n-2}$.
Comparing coefficients of $t^{n-1}$ yields:
$n = -\frac{3}{2}$.

Detailed analysis of options:

• 0 is incorrect.
• is incorrect.
• $-1/2$ corresponds to 1D diffusion, not 3D spherical coordinates.
• $-3/2$ is the correct value corresponding to 3-dimensional spherical diffusion.

The given differential equation is of the form $M\,dx + N\,dy = 0$ with:
$M = \cos(x+y)$,
$N = 3y^2 + 2y + \cos(x+y)$.
Check for exactness:
$\frac{\partial M}{\partial y} = -\sin(x+y)$,
$\frac{\partial N}{\partial x} = -\sin(x+y)$.
Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the equation is exact.
The solution is obtained by:
$\int M\,dx$ (treating $y$ as constant) $+ \int (\text{terms in } N \text{ free from } x)\,dy = C$.
$\int \cos(x+y)\,dx = \sin(x+y)$.
Terms of $N$ free from $x$: $3y^2 + 2y$, so $\int (3y^2 + 2y)\,dy = y^3 + y^2$.
Combining gives:
$\sin(x+y) + y^3 + y^2 = C$.

Detailed analysis of options:

• $\sin(x+y) + y^3 + y^2 = C$ is correct.
• Has $x-y$ instead of $x+y$.
• Replaces $y^3 + y^2$ with $x^3 + x^2$.
• Has $x-y$ and $x$ terms.