Guillaume François Antoine, Marquis de L’Hospital (1661 ? 2 February 1704), was a French mamatician. His name is firmly associated with L’Hospital’s rule for calculating limits involving indeterminate forms 0/0 and ?/?.

\(u(x,t) = X(x) T(t) \)
\(x^2 X’Y+y^2 Y’X=XY\)
\(X = ce^{\frac{-k}{x}} \)
\(Y = c?e^{\frac{k-1}{y}} \)
\(u(x,t) = cc? e^{\frac{-k}{x}} e^{\frac{k-1}{y}} \)
\(u(0,y) = e^{\frac{2}{y}}= cc? e^{\frac{k-1}{y}} \)
k = 3 and cc? = 1
Therefore \(u(x,t) = e^{\frac{-3}{x}} e^{\frac{2}{y}}. \)

The heat equation is used is several fields, some of which include,

• Probability ory (to study Brownian motion)
• Financial Mamatics (for solving Black- Scholes PDE)
• Quantum Mechanics (to determine spread of wave function)

0 and 30 are end points. So, at se points function is zero. Hence u(0,t) = 0 = u(30,t). Next due to steady state conditions, at beginning that is initial conditions, we have u(x,0) = 20 + 60/30 x.

If we have a function u(x,t), n function u depends on both x and t. For using variable separable method we assume that it can be written as u(x,t) = X(x).T(t) where X depends only on x and T depends only on t.

When solving a partial differential equation using a variable separable method, n function can be written as product of functions depending on one variable only.

The order of an equation is defined as highest derivative present in equation. Hence, in given equation, \(\frac{?z}{?x}-(\frac{?z}{?y})^5=0,\) order is 1.

If A is n × n, B is m × m and I_k denotes k × k identity matrix n we can define what is sometimes called Kronecker sum, ?, by,
A?B=A?I_M+B?I_N

Since this is a heat equation, solution must be a transient solution, that is it should decay as time increases. This happens only when k is negative and solution comes out to be (c??cospx + c???sinpx) (c e^{-c2 p2 t}).

\(u(x,t) = X(x) T(t) \)
\(x^3 X’Y+y^2 Y’X=0\)
\(\frac{X?}{ X} = \frac{k}{x^3} \) which implies \(X = ce^{\frac{k}{2x^2}}\)
\(\frac{Y?}{Y} = \frac{-k}{y^2} \) which implies \(Y = c? e^{\frac{k}{y}}\)
\(u(x,t) = cc?e^{\frac{k}{2x^2}} e^{\frac{k}{y}} \)
\(u(0,y) = 10e^{\frac{5}{y}}= cc?e^{\frac{k}{y}} \)
Therefore k = 5 and cc? = 10
Hence, \(u(x,t) = 10e^{\frac{-5}{2x^2}} e^{\frac{5}{y}}. \)