Differential Equation

When we are given a derivative like \(\frac{dy}{dx} = x^2\), we can easily find the original function \(y\) as the indefinite integral of \(x^2\) : \(y = \int x^2 dx = \frac{x^3}{3} + C\).

However, if the derivative is not a simple function of only \(x\), for example, if \(\frac{dy}{dx} = x^2 e^y\),

• how can we represent \(y\) as an explicit function of \(x\) (find a function \(f(x)\) such that \(y = f(x)\))?
• if we can't find such \(f\), to what extent can we simplify the relationship between \(x\) and \(y\)? For example, can we represent the relationship between \(x\) and \(y\) without using derivative?

In this section, we will partly answer this question.

As we saw above, from \(\frac{dy}{dx} = x^2\) we get \(y = \int x^2 dx = \frac{x^3}{3} + C\), which contains an arbitrary constant \(C\). It is a family of solutions, rather than a single solution, so it is called the General Solution.

However, if we are given particular initial values of \(x\) and \(y\), ie, \((x,y) = (x_0, y_0)\), then we can calculate a unique value of \(C\) by replacing \(x\) and \(y\) with \(x_0\) and \(y_0\) respectively in the general solution. In the above example, if the initial values are \((x,y)=(1,2)\), then by replacing \(x\) and \(y\) with \(1\) and \(2\) in the general solution \(y = \int x^2 dx = \frac{x^3}{3} + C\), we get \(2 = \frac{1^3}{3} + C \implies C = \frac{5}{3}\). Put \(C = \frac{5}{3}\) into the general solution, and we get \(y = \frac{x^3}{3} + \frac{5}{3}\). This is just a single solution, and it is called the particular solution corresponding to initial values \((x,y)=(1,2)\).

The general solution of the differential equation \(\frac{dy}{dx} = g(x)\) is \(y = \int g(x) dx\).

Find \(y\) if \(\frac{dy}{dx} - 3x^2 - e^{2x} = \frac{5}{x}\). Find a particular solution for initial value \(x = 1, y = 2\).

• General Solution

  From \(\frac{dy}{dx} - 3x^2 - e^{2x} = \frac{5}{x}\) we get \(\frac{dy}{dx} = 3x^2 + e^{2x} + \frac{5}{x} = g(x)\).

  Hence, by formula, the general solution is

  \(y = \int g(x) dx = \int \left( 3x^2 + e^{2x} + \frac{5}{x} \right) dx\)

  \(= \int 3x^2 dx + \int e^{2x} dx + \int \frac{5}{x} dx\)

  \(= x^3 + \frac{e^{2x}}{2} + 5ln|x| + C\)

• Particular Solution

  Putting \(x = 1, y = 2\) into the general solution, we have \(2 = 1^3 + \frac{e^{2 \cdot 1}}{2} + 5ln|1| + C\) and thus \(C = 1 - \frac{e^2}{2}\).

  Putting \(C = 1 - \frac{e^2}{2}\) into the general solution, we get the particular solution for \(x = 1, y = 2\) :

  \(y = x^3 + \frac{e^{2x}}{2} + 5ln|x| + 1 - \frac{e^2}{2}\)

The general solution of the separable differential equation \(\frac{dy}{dx} = \frac{p(x)}{q(y)}\) is \(\int q(y) dy = \int p(x) dx\).

Find the general solution for \(\frac{dy}{dx} - (2y - 1)^2 = 0\) and find the particular solution for initial value \(x = 3, y = 1\).

• General Solution

  The equation can be rewritten as \(\frac{dy}{dx} = (2y - 1)^2 = \frac{1}{(2y - 1)^{-2}} = \frac{p(x)}{q(y)}\) where

  • \(p(x) = 1\)
  • \(q(y) = (2y - 1)^{-2}\)

  so the general solution is \(\int q(y) dy = \int p(x) dx\).

  First, we need find \(\int q(y) dy\).

  According to our experience, integration by substitution should be used. Let \(u = u(y) = 2y - 1\). Then we have \(du = u' dy = 2 dy\) and thus

  \(\int q(y) dy = \int (2y - 1)^{-2} dy\)

  \(= \int (2y - 1)^{-2} \cdot \frac{2}{2} dy\)

  \(= \frac{1}{2} \int u^{-2} du\)

  \(= \frac{1}{2} \cdot \frac{u^{-2+1}}{-2+1} + C_1\)

  \(= \frac{u^{-1}}{-2} + C_1\)

  \(= \frac{1}{-2u} + C_1\)

  \(= \frac{1}{-2(2y - 1)} + C_1\)

  Second, we find \(\int p(x) dx = \int 1 dx = x + C_2\).

  Finally, the general solution is \(\frac{1}{-2(2y - 1)} + C_1 = x + C_2 \implies \frac{1}{-2(2y - 1)} - x + C = 0\).

  Optionally (not required for your homework, quiz or exam), we can simplify the answer and write \(y\) as an explicit function of \(x\) :

  \(\frac{1}{-2(2y - 1)} + C_1 = x + C_2\)

  \(\implies \frac{1}{-2(2y - 1)} = x + C\)

  \(\implies -2(2y - 1) = \frac{1}{x+C}\)

  \(\implies 2y - 1 = \frac{1}{-2(x+C)}\)

  \(\implies 2y = \frac{1}{-2(x+C)} + 1\)

  \(\implies y = \frac{1}{-4(x+C)} + \frac{1}{2}\)

• Particular Solution

  Putting \(x = 3, y = 1\) into the general solution, we have \(1 = \frac{1}{-4(3+C)} + \frac{1}{2}\) and thus \(C = \frac{-7}{2}\).

  Putting \(C = \frac{-7}{2}\) into the general solution, we get the particular solution for \(x = 3, y = 1\) :

  \(y = \frac{1}{-4(x-\frac{7}{2})} + \frac{1}{2}\)

Find the general solution for \(\frac{dy}{dx} = \frac{y(2x^2 + 3x + 1)}{(3y^3 + 4y)x}\) and find the particular solution for initial value \(x = 1, y = 1\).

• General Solution

  The equation can be rewritten as

  \(\frac{dy}{dx} = \frac{y(2x^2 + 3x + 1)}{(3y^3 + 4)x}\) \(= \frac{2x^2 + 3x + 1}{x} \cdot \frac{y}{3y^3 + 4}\) \(= \frac{\frac{2x^2 + 3x + 1}{x}}{\frac{3y^3 + 4}{y}}\) \(= \frac {2x + 3 + \frac{1}{x}} {3y^2 + \frac{4}{y}}\) \(= \frac {p(x)}{q(y)}\)

  where

  • \(p(x) = 2x + 3 + \frac{1}{x}\)
  • \(q(y) = 3y^2 + \frac{4}{y}\)

  so the general solution is \(\int q(y) dy = \int p(x) dx\).

  First, we find

  \(\int q(y) dy = \int 3y^2 + \frac{4}{y} dy\)

  \(= \int 3y^2 dy + \int \frac{4}{y} dy\)

  \(= y^3 + 4ln|y| + C_1\)

  Second, we find

  \(\int p(x) dx = \int 2x + 3 + \frac{1}{x} dx\)

  \(= \int 2x dx + \int 3 dx + \int \frac{1}{x} dx\)

  \(= x^2 + 3x + ln|x| + C_2\)

  Finally, the general solution is \(y^3 + 4ln|y| + C_1 = x^2 + 3x + ln|x| + C_2\), or equivalently

  \(y^3 + 4ln|y| - x^2 - 3x - ln|x| + C = 0\)

  In this example, it is quite hard (if not impossible) for us to represent \(y\) as an explicit function of \(x\).

• Particular Solution

  Putting \(x = 1, y = 1\) into the general solution, we have \(1^3 + 4ln|1| - 1^2 - 3 \cdot 1 - ln|1| + C = 0\) and thus \(C = 3\).

  Putting \(C = 3\) into the general solution, we get the particular solution for \(x = 1, y = 1\) :

  \(y^3 + 4ln|y| - x^2 - 3x - ln|x| + 3 = 0\).

Find the general solution for \(\frac{dy}{dx} = \frac{x(y^3+1)^4}{y^2 e^x}\) and find the particular solution for initial value \(x = 1, y = 1\).

• General Solution

  The equation can be rewritten as

  \(\frac{dy}{dx} = \frac{x(y^3+1)^4}{y^2 e^x}\) \(= \frac{x}{e^x} \cdot \frac{(y^3+1)^4}{y^2}\) \(= x e^{-x} \cdot \frac{1}{\frac{y^2}{(y^3+1)^4}}\) \(= \frac{x e^{-x}}{\frac{y^2}{(y^3+1)^4}}\) \(= \frac {p(x)}{q(y)}\)

  where

  • \(p(x) = x e^{-x}\)
  • \(q(y) = \frac{y^2}{(y^3+1)^4}\)

  so the general solution is \(\int q(y) dy = \int p(x) dx\).

  First, we find \(\int q(y) dy = \int \frac{y^2}{(y^3+1)^4} dy\) by substitution.

  Let \(u=y^3+1\). Then \(du = u'dy = 3y^2 dy\) and

  \(\int q(y) dy = \int \frac{y^2}{(y^3+1)^4} dy\)

  \(= \int \frac{3y^2}{3(y^3+1)^4} dy\)

  \(= \frac{1}{3}\int \frac{1}{(y^3+1)^4} \cdot 3y^2dy\)

  \(= \frac{1}{3}\int \frac{1}{u^4} du\)

  \(= \frac{1}{3}\int u^{-4} du\)

  \(= \frac{1}{3} \left( \frac{u^{-4+1}}{-4+1} + C \right)\)

  \(= \frac{1}{3} \left( \frac{u^{-3}}{-3} + C \right)\)

  \(= \frac{u^{-3}}{-9} + C_1\)

  \(= \frac{(y^3+1)^{-3}}{-9} + C_1\)

  Second, we find \(\int p(x) dx = \int x e^{-x} dx\) by integration by parts.

  Let \(u = x\) and \(v' = e^{-x}\). Then \(u' = 1\), \(v = \int e^{-x} dx = \frac{e^{-x}}{-1} + C = -e^{-x} + C\). Without loss of generality, let \(C=0\) and then \(v = -e^{-x}\) and

  \(\int p(x) dx = \int x e^{-x} dx = \int u v' dx\)

  \(= uv - \int u'v dx\)

  \(= x \cdot (- e^{-x}) - \int 1 \cdot (- e^{-x}) dx\)

  \(= - xe^{-x} + \int e^{-x} dx\)

  \(= - xe^{-x} + \frac{e^{-x}}{-1} + C_2\)

  \(= - xe^{-x} - e^{-x} + C_2\)

  Finally, the general solution is \(\frac{(y^3+1)^{-3}}{-9} + C_1 = - xe^{-x} - e^{-x} + C_2\), or equivalently

  \(\frac{(y^3+1)^{-3}}{-9} + xe^{-x} + e^{-x} + C = 0\)

  In this example, \(y\) can be written as an explicit function of \(x\). Do it by yourself (not required, though).

• Particular Solution

  Putting \(x = 1, y = 1\) into the general solution, we have \(\frac{(1^3+1)^{-3}}{-9} + 1 \cdot e^{-1} + e^{-1} + C = 0\) and thus \(C = \frac{1}{72} - \frac{2}{e}\).

  Putting \(C = \frac{1}{72} - \frac{2}{e}\) into the general solution, we get the particular solution for \(x = 1, y = 1\) :

  \(\frac{(y^3+1)^{-3}}{-9} + xe^{-x} + e^{-x} + \frac{1}{72} - \frac{2}{e} = 0\).

A linear first-order differential equation is an equation of the form \(\frac{dy}{dx} + P(x) \cdot y = Q(x)\)

Remark:

• "Linear" : If we move the term \(P(x) \cdot y\) to the RHS of the equation, then we get \(\frac{dy}{dx} = Q(x) - P(x) \cdot y\), ie, the derivative \(\frac{dy}{dx}\) is a linear function of \(y\) if we regard \(x\) as a constant.
• "First-Order" : only the first order derivative exists in the equation, no second derivative or higher-order derivative.

The function \(I(x) = e^{\int P(x) dx}\) is called an integrating factor for the linear first-order differential equation \(\frac{dy}{dx} + P(x) \cdot y = Q(x)\).

Remark: \(\int P(x) dx\), as an indefinite integral, is a family of antiderivatives indexed by an arbitrary constant \(C\). For example, if \(P(x) = 2x\), then \(\int P(x) dx = \int 2x dx = x^2 + C\). However, for finding the solution to a linear first-order differential equation, one special antiderivative of \(P(x)\) is enough. For example, for \(P(x) = 2x\) and thus \(\int P(x) dx = x^2 + C\), we can choose the special antiderivative \(x^2\) corresponding to \(C=0\), and then get \(I(x) = e^{x^2}\), instead of the full version \(I(x) = e^{x^2 + C}\). Why is it enough to use just \(I(x) = e^{x^2}\) rather than the full version \(I(x) = e^{x^2 + C}\)? Actually, you can try both \(I(x) = e^{x^2}\) and \(I(x) = e^{x^2 + C}\) in the following procedure for solving linear first-order differential equations, and you will eventually find that these two different versions of \(I(x)\) yield exactly the same final solution. Therefore, for convenience, we can just ignore the constant \(C\) in \(\int P(x) dx\).

1. Write the equation in the standard form \(\frac{dy}{dx} + P(x) \cdot y = Q(x)\).
2. Find the integrating factor \(I(x) = e^{\int P(x) dx}\). One special antiderivative is enough, so we can drop the arbitrary constant \(C\).
3. Calculate \(\int Q(x) \cdot I(x) dx\), which is a family of antiderivatives of \(Q(x)I(x)\)
4. The general solution is \(y = \frac{\int Q(x) \cdot I(x) dx}{I(x)}\)

Remark: You are not required to know the derivation of the solution. You can verify the solution by yourself : replace \(y\) with \(\frac{\int Q(x) \cdot I(x) dx}{I(x)}\) in the original different equation, and you will find the LHS and RHS are equal, which means it is truely a solution to the equation.

Suppose that today you open a new bank account and deposit \(y_0\) dollars (called principal) into it. Denote by \(y=y(x)\) the total amount of money you will have after time \(x\). If \(r\) is the annual interest rate compounded continuously, then the rate of growth of your total money is \[ \frac{dy}{dx} = ry \]

Suppose that after the initial principal \(y_0\) you continue depositing into your account continuous money at a constant rate \(D\) dollars/year. Then the eventual rate of growth of your total money is \[ \boxed{ \frac{dy}{dx} = ry + D } \]

Remark: In the textbook, the symbols \(t\) and \(A\) are used to represent "time" and "total amount" respectively, but here I use \(x\) and \(y\), because many people may regard \(t\) and \(A\) as constants rather than variables.

The above is a general solution to the differential equation \(\frac{dy}{dx} = ry + D\). The general solution is a class of functions indexed by the arbitrary constant \(C\). Obviously, if \(y_0\), \(r\) and \(D\) are all given as specific numbers, then at a specific time \(x\), the total amount of money you will have should be a determinate number, rather than an arbitrary quantity. That means, we should be able to figure out the value of \(C\). As we did before, if an initial condition is given, then we can find a unique solution to the differential equation, which is call the particular solution.

The initial condition for the continuous deposit problem is \(x=0, y=y_0\), because at the initial time \(x=0\) the total amount of money is \(y=y_0\). Then we can find the corrsponding particular solution:

• Replace \(x\) and \(y\) with \(0\) and \(y_0\) respectively in the general solution, and we get

  \(y_0 = \frac{C e^{r\cdot 0} - D}{r} \implies y_0 = \frac{C - D}{r} \implies C = ry_0 + D\)

• Replace \(C\) with \(ry_0+D\) in the general solution, and we get the particular solution for \(x=0, y=y_0\):

  \[ \boxed{y = \frac{(ry_0+D)e^{rx} - D}{r} }\]

Note that the particular solution \(y = y(x) = \frac{(ry_0+D)e^{rx} - D}{r}\) means that the total amount of money \(y\) is a function of time \(x\). At a specific time \(x=x_t\), your total amount of money \(y\) can be calculated by replacing \(x\) with \(x_t\) in the particular solution, ie,

\[ \boxed{y(x_t) = \frac{(ry_0+D)e^{rx_t} - D}{r} }\]

Suppose today you deposit \(10000\) dollars in an IRA at \(3\%\) interest compounded continuously. In addition, you make continuous deposit at the rate of \(1200\) dollars per year. How much will you have in total after 10 years?

We identity \(y_0 = 10000\), \(r = 3\% = 0.03\), \(D = 1200\), and \(x_t=10\).

By the formula, after 10 years, you will have \(y(10) = \frac{(ry_0+D)e^{rx_t} - D}{r} = \frac{(0.03\cdot 10000 + 1200)e^{0.03\cdot 10} - 1200}{0.03} = 27492.9\) dollars.