10. Differential equations

Differential equations

We consider equations in which there appear a function x(t), its argument t and its derivative x'. Namely, equations of form

    \[x'=g(x,t),\]

where g is a function of two variables, e.g.:

    \[x'=x^2t.\]

A solution to such an equation is a function x(t), such that its derivative actually can be written in this form, i.e in this case x'(t)=x^2t. In on the right hand side (in g) there is no x, this is simply an integral. If not, it becomes more complicated. But still we can verify that any function of form x(t)=\frac{1}{-t^2/2+C}, where C is any constant, satisfies the equation. Indeed,

    \[x'=\frac{t}{(-t^2/2+C)^2}=tx^2.\]

But how to come up with such a function? Generally it is hard to do so, but for some forms of function g it is possible to do it quite easily. We describe few such cases below.

Notice that the solution is usually determined up to an arbitrary constant. By adding an additional condition, called a boundary condition, to a differential equation, we can determine such a constant (this is called Chauchy problem), e.g.

    \[\begin{cases}x'=x^2t,\\ x(1)=1\end{cases}\]

we get (since we know the solution to the differential equation) x(1)=\frac{1}{(-1/2+C)}=1, so C=3/2, thus in this case x(t)=\frac{1}{-t^2/2+3/2}.

Separable equation

An equation x'=g(x,t) is separable if g(x,t)=a(x)b(t). Then x'/a(x)=b(t), so \int b(t)\, dt=\int \frac{x'(t)}{a(x(t))}\,dt= \int \frac{1}{a(x)}\,dx. This gives an equation with x and t only, so we can deduce x(t) out of it.

For example the equation

    \[x'=x^2t.\]

We get

    \[\int \frac{dx}{x^2}=\int t\,dt,\]

so -1/x+C=t^2/2, thus x(t)=1/(C-t^2/2).

Homogeneous equation

An equation x'=g(x,t) is homogeneous, if g(x,t)=f(x/t). The substitution z(t)=\frac{x(t)}{t} proves to be useful. We get x(t)=tz(t), so x'(t)=z(t)+tz'(t), thus the equation takes the form f(z)=z+tz', so we get z'=\frac{1}{t}\cdot (f(z)-z), which is a separable equation which we already know how to solve.

Linear equation

An equation x'=g(x,t) is linear if g(x,t)=a(t)x+b(t). Such an equation can be solved by solving first an linear homogeneous equation x'=a(t)x, which is simply a separable equation with solution x_0(t)=Ce^{\int a(t)\,dt}. Notice that the additional term +b(t) in the derivative of x can be achieved, if C is a function C(t) rather than a constant. Thus, we are looking for solutions of form x(t)=C(t)e^{A(t)}, where A(t)=\int a(t)\, dt. On one hand, x'=C'e^{A(t)}+Ca(t)e^{A(t)}, but on the second hand a(t)x+b(t)=Ca(t)e^{A(t)}+b(t), so C'(t)e^{A(t)}=b(t), hence C'(t)=b(t)e^{-A(t)}, thus C(t)=\int b(t)e^{-A(t)}\,dt, which can be substituted into x(t)=C(t)e^{A(t)} to get the final solution.

Bernoulli equation

An equation x'=g(x,t) is called Bernoulli equation, if g(x,t)=a(t)x+b(t)x^B, where B\in\mathbb{R}\setminus\{0,1\}. In this case, if we substitute u(t)=x(t)^{-B+1} we get u'=(-B+1)x'x^{-B}, so u'/(1-B)=x'x^{-B}, thus since x'= a(t)x+b(t)x^B, we get u'/(1-B)=x'x^{-B}=ax^{1-B}+b=au+b, which is a linear equation for u, which we already know how to solve.