Question

Solve the equation dt2dx2−4t=0

Answer 1

Answer on Question #69640 – Math – Differential Equation

Question

Solve the equation dt2dx2-4t=0

Solution

Case 1. If the differential equation is separable, then

$$\left(\frac{dt(x)}{dx}\right)^2 - 4t(x) = 0$$

Solve for $\frac{dt(x)}{dx}$

$$\frac{dt(x)}{dx} = -2\sqrt{t(x)} \quad \text{or} \quad \frac{dt(x)}{dx} = 2\sqrt{t(x)}$$

For $\frac{dt(x)}{dx} = -2\sqrt{t(x)}$

Divide both sides by $2\sqrt{t(x)}$ and multiply by $dx$ :

$$\frac{dt(x)}{2\sqrt{t(x)}} = -dx$$

Integrate both sides:

$$\int \frac{dt(x)}{2\sqrt{t(x)}} = -\int dx + c_1, \text{ where } c_1 \text{ is an integration constant.}$$

Solve for $t(x)$ :

$$t(x) = (-x + c_1)^2$$

For $\frac{dt(x)}{dx} = 2\sqrt{t(x)}$

Divide both sides by: $2\sqrt{t(x)}$ and multiply by $dx$ :

$$\frac{dt(x)}{2\sqrt{t(x)}} = dx$$

Integrate both sides:

$$\int \frac{dt(x)}{2\sqrt{t(x)}} dx = \int dx + c_1, \text{ where } c_1 \text{ is an integration constant.}$$

Solve for $t(x)$ :

$$t(x) = (x + c_1)^2$$

Answer: $t(x) = (-x + c_1)^2$ or $t(x) = (x + c_1)^2$ .

Case 2. If it is a second order linear differential equation

$$\frac {d ^ {2} t}{d x ^ {2}} - 4 t = 0,$$

then

$$\lambda^ {2} - 4 = 0,$$ $$\lambda_ {1} = 2, \lambda_ {2} = - 2,$$ $$t (x) = c _ {1} e ^ {\lambda_ {1} x} + c _ {2} e ^ {\lambda_ {2} x} = c _ {1} e ^ {- 2 x} + c _ {2} e ^ {2 x},$$

where $c_{1}$ and $c_{2}$ are arbitrary real constants.

Answer: $t(x) = c_1e^{-2x} + c_2e^{2x}$.

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