Answer to Question #207676 in Differential Equations for Mitchy

Question

solve the intergrating factt ;tdt+xdx=a2(tdx-xdx)/t^2+x^2

Answer 1

Solve the integrating factor:

$tdt+xdx=\frac{a^2(tdx-xdt)}{t^2+x^2}$

Solution:

$(x-\frac{a^2t}{t^2+x^2})dx+(t+\frac{a^2x}{t^2+x^2})dt=0$

$Pdx+Qdt=0$

where $P=(x-\frac{a^2t}{t^2+x^2})$ and $Q=(t+\frac{a^2x}{t^2+x^2})$ .

Let's find $\frac{\partial P}{\partial t}$ and $\frac{\partial Q}{\partial x}$:

$\frac{\partial P}{\partial t}=-a^2\cdot\frac{x^2+t^2-2t^2}{(t^2+x^2)^2}=a^2\cdot\frac{t^2-x^2}{(t^2+x^2)^2}$;

$\frac{\partial Q}{\partial x}=a^2\cdot\frac{t^2+x^2-2x^2}{(t^2+x^2)^2}=a^2\cdot\frac{t^2-x^2}{(t^2+x^2)^2}$ .

Since $\frac{\partial P}{\partial t}=\frac{\partial Q}{\partial x}$ than we can conclude that it is a total differential equation.

Its solution can be find as:

$U(x,t)=\int_{x_0}^x P(\tau,t_0)d\tau+\int_{t_0}^t Q(x,\tau)d\tau$

where $x_0$ and $t_0$ are any values that can be freely chosen. Let's choose $x_0=1$, $t_0=0$ .

$U(x,t)=\int_{1}^x (\tau-\frac{a^2\cdot0}{0^2+\tau^2})d\tau+\int_{0}^t (\tau+\frac{a^2x}{\tau^2+x^2})d\tau=$

$\int_{1}^x \tau d\tau+\int_{0}^t (\tau+\frac{a^2x}{\tau^2+x^2})d\tau=$ $\frac{x^2}{2}-\frac12+\frac{t^2}{2}+a^2\arctan{\frac {t}{x}}$ .

So the general integral of the equation has the form:

$\frac{x^2}{2}+\frac{t^2}{2}+a^2\arctan{\frac {t}{x}}=const$ .

Answer:

This equation is a total differential equation therefore, its solution can be found without an integrating factor or any constant value can be used as integrating factor in this case.

Solution: $\frac{x^2}{2}+\frac{t^2}{2}+a^2\arctan{\frac {t}{x}}=const$ .