Before we move on to the next section, make sure you can solve the following problems analytically using pen and paper. Hints and solutions are available.
Solve $\frac{dy}{dx} = x^2$ subject to $y(0) = 1$.
Try just integrating the right-hand side.
Integrating both sides yields
$$ y = \int x^2\,dx + C \,. $$
This is the general solution to the differential equation, to which we now need to apply the initial condition $y(0)=1$.
$ 1 = y(0) = C$, so we take $C=1$. The solution is therefore
$$ y = \frac{x^3}{3} + 1 \,. $$
Solve $\frac{dy}{dx} = \frac{x^2}{y}$, suvject to $y(0) = 1$.
Try separation of variables.
Separation of variables yields
$$ \int y\,dy = \int x^2\,dx + C \,. $$
Therefore
$$ \frac{y^2}{2} = \frac{x^3}{3} + C\,. $$
On rearranging we get
$$ y = \pm \sqrt{2\left(\frac{x^3}{3} + C\right)} $$
This is the general solution to the differential equation, to which we now need to apply the initial condition $y(0)=1$.
$1=y(0) = \pm \sqrt{0+C}$, so we take the positive root and $C=1$. The solution is therefore
$$ y = \sqrt{\frac{2x^3}{3} + 1} \,.$$
Solve $\frac{dy}{dx} + \frac{2y}{x} = x^4$ subject to $y(1) = 1$.
Try using an integrating factor.
The integrating factor is
$$ \varphi = e^{\int \frac{2}{x}\,dx} = x^2 \,. $$
The general solution to the equation is therefore
$$ y = \frac{x^5}{7} + \frac{C}{x^2} \,. $$
Applying the initial condition gives
$$ 1 = y(1) = \frac{1}{7} + C \,, $$
so $C = \frac{6}{7} $. The solution is therefore
$$ y = \frac{x^5}{7} + \frac{6}{7x^2} \,.$$
Solve $\frac{dy_1}{dx} = -y_2$ and $\frac{dy_2}{dx} = y_1$, subject to $y_1(0) = 1$ and $y_2(0) = 0$.
Combining these two equations, by differentiating the first and substituting it in the second, yields a second-order ODE with linear coefficients. Look at the characteristic equation.
Combining these two equations (differentiating the first and substituting it into the second) gives
$$ \frac{d^2y_1}{dx^2}=−y_1 \,. $$
The general solution to this is
$$ y_1=A\cos(x)+B\sin(x) \,. $$
From the first equation we can calculate $y_2$ as
$$ y_2=A\sin(x) − B\cos(x)\,.$$
Applying the initial conditions gives $A=1$ and $B=0$.
The solution is therefore
$$ y_1=\cos(x) \quad \textrm{and} \quad y_2=\sin(x). $$
Solve $\frac{d^2y}{dx^2} + 3\frac{dy}{dx} - 4y = 0$, subject to $y(0)=1$ and $y'(0) = 0$.
This is a second-order ODE with linear coefficients. Look at the characteristic equation.
The general solution to this is
$$ y=Ae^x+Be^{−4x} \,. $$
Applying the initial conditions gives $A=\frac{4}{5}$ and $B=\frac{1}{5}$.
The solution is therefore
$$ y=\frac{4}{5}e^x+\frac{1}{5}e^{−4x} \,.$$
There are many methods for solving ODEs analytically, but we may not be able to calculate explicit forms for the integrals, even if we can formulate a given ODE into one of the known forms. Furthermore, if we wish to solve an equation that does match one of the standard forms, then we must recourse to numerical methods. This is the case for most systems of ODEs that represent real-world situations, since most of these are highly non-linear.