Hw1 p4

Problem 4

(a)

If equation (1) is exact, there's a function \(H(x,t)\) such that its total derivative with respect to \(t\) is the LHS of (1). Using the multivariable chain rule, the total derivative of \(H(x(t), t)\) is: $$ \frac{d}{dt}H(x,t) = \frac{\partial H}{\partial t} + \frac{\partial H}{\partial x}\frac{dx}{dt} $$ Compare to equation (1): $$ f(x,t) = \frac{\partial H}{\partial t} \equiv H_t(x,t) $$ $$ g(x,t) = \frac{\partial H}{\partial x} \equiv H_x(x,t) $$ The problem assumes that \(f(x,t)\) and \(g(x,t)\) have continuous first-order partial derivatives. So \(H(x,t)\) has continuous second-order partial derivatives.

By Clairaut's Theorem on the equality of mixed partial derivatives: $$ \frac{\partial^2 H}{\partial x \partial t} = \frac{\partial^2 H}{\partial t \partial x} \quad \text{or} \quad H_{tx} = H_{xt} $$ Partial derivates \(f\) and \(g\): $$ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( \frac{\partial H}{\partial t} \right) = H_{tx} $$ $$ \frac{\partial g}{\partial t} = \frac{\partial}{\partial t} \left( \frac{\partial H}{\partial x} \right) = H_{xt} $$ \(H_{tx} = H_{xt}\), so: $$ f_x = g_t $$ This completes the proof.

(b)

i. \(4t^3x^3 + 3t^2 + (3t^4x^2 + 6x^5)\frac{dx}{dt} = 0\)

CHeck exactness: The ODE is form \(f(x,t) + g(x,t)\frac{dx}{dt} = 0\) with: \(f(x,t) = 4t^3x^3 + 3t^2\) \(g(x,t) = 3t^4x^2 + 6x^5\)

$$ f_x = \frac{\partial}{\partial x}(4t^3x^3 + 3t^2) = 12t^3x^2 $$ $$ g_t = \frac{\partial}{\partial t}(3t^4x^2 + 6x^5) = 12t^3x^2 $$ Since \(f_x = g_t\), the equation is exact.

Solve the Equation: Find a potential function \(H(x,t)\) such that \(H_t = f\) and \(H_x = g\). $$ H(x,t) = \int (3t^4x^2 + 6x^5) \, dx = 3t^4\left(\frac{x^3}{3}\right) + 6\left(\frac{x^6}{6}\right) + \psi(t) $$ $$ H(x,t) = t^4x^3 + x^6 + \psi(t) $$ where \(\psi(t)\) is an arbitrary function of \(t\). $$ H_t = \frac{\partial}{\partial t}(t^4x^3 + x^6 + \psi(t)) = 4t^3x^3 + \psi'(t) $$ $$ H_t = f(x,t) \implies 4t^3x^3 + \psi'(t) = 4t^3x^3 + 3t^2 $$ $$ \psi'(t) = 3t^2 $$ $$ \psi(t) = \int 3t^2 \, dt = t^3 $$ (Absorb the constant of integration into the final solution's constant). The potential function is \(H(x,t) = t^4x^3 + x^6 + t^3\). The solution to an exact ODE is given by \(H(x,t) = C\), where \(C\) is a constant.

Solution is $$ t^4x^3 + x^6 + t^3 = C $$

ii. \(t(\cot x) \frac{dx}{dt} = -2\)

Check exactness: Rewrite the ODE in the standard form \(f(x,t) + g(x,t)\frac{dx}{dt} = 0\): $$ 2 + (t \cot x)\frac{dx}{dt} = 0 $$ \(f(x,t) = 2\) \(g(x,t) = t \cot x\)

\[ f_x = \frac{\partial}{\partial x}(2) = 0 $$ $$ g_t = \frac{\partial}{\partial t}(t \cot x) = \cot x $$ Since $f_x \neq g_t$, the equation as written is not exact. But professor gave hint, this equation needs an integrating factor to become exact. Multiplying integrating factor $\mu(t,x) = t\sin x$ gives an exact equation: $$ (t\sin x) \left( 2 + t \cot x \frac{dx}{dt} \right) = 0 $$ $$ 2t\sin x + t^2 \sin x \frac{\cos x}{\sin x} \frac{dx}{dt} = 0 $$ $$ 2t\sin x + t^2 \cos x \frac{dx}{dt} = 0 \]

Excact ODE: \(2t\sin x + t^2 \cos x \frac{dx}{dt} = 0\). \(f(x,t) = 2t\sin x\) \(g(x,t) = t^2 \cos x\)

Check for exactness again: $$ f_x = \frac{\partial}{\partial x}(2t\sin x) = 2t\cos x $$ $$ g_t = \frac{\partial}{\partial t}(t^2 \cos x) = 2t\cos x $$ Since \(f_x = g_t\), the new equation is exact. Find a potential function \(H(x,t)\) where \(H_t = f\) and \(H_x = g\). $$ H(x,t) = \int 2t\sin x \, dt = t^2 \sin x + \phi(x) $$ where \(\phi(x)\) is an arbitrary function of \(x\). $$ H_x = \frac{\partial}{\partial x}(t^2\sin x + \phi(x)) = t^2\cos x + \phi'(x) $$ $$ H_x = g(x,t) \implies t^2\cos x + \phi'(x) = t^2\cos x $$ $$ \phi'(x) = 0 \implies \phi(x) = K \quad (\text{a constant}) $$ The potential function is \(H(x,t) = t^2\sin x + K\). The solution is given by \(H(x,t) = C_1\).

Solution: $$ t^2\sin x = C $$ (where \(C = C_1 - K\)).