# To find particular solutions

Up a level : Differential Equations
Previous page : Separable differential equations - A few illustrating examples
Next page : Homogenous functions The general solution to a differential equation will contain unknown constants. We for example found that $s'' = a$

has the solution $s = \frac{{a{t^2}}}{2} + {v_0}t + {s_0}$

To start with we have a given constant, a, then we have two constants we may need to find. That is the initial velocity and the initial position, and those two might be given. In that case we say we have given initial conditions. We might also be given other conditions, like two points. Say we have that after 1 s we are at 10 m and at 5 s we are at 50 m, and that the acceleration is -10 m/s2. We have $s = \frac{{ - 10{t^2}}}{2} + {v_0}t + {s_0} = - 5{t^2} + {v_0}t + {s_0}$

And the two points give us $\left\{ \begin{gathered} 10 = - 5 \cdot {1^2} + {v_0} \cdot 1 + {s_0} \hfill \\ 50 = - 5 \cdot {5^2} + {v_0} \cdot 5 + {s_0} \hfill \\ \end{gathered} \right.$

or $\left\{ \begin{gathered} 10 = - 5 + {v_0} + {s_0} \hfill \\ 50 = - 125 + 5{v_0} + {s_0} \hfill \\ \end{gathered} \right.$

That gives us $\left\{ \begin{gathered} 15 = {v_0} + {s_0} \hfill \\ 175 = 5{v_0} + {s_0} \hfill \\ \end{gathered} \right.$

Solving this gives us ${v_0} = 40{\text{ m/s}},\quad {s_0} = - 25{\text{ m}}$

We thus get the particular solution $s = - 5{t^2} + 40t - 25$

In general first order differential equations will have one constant of integration, a second order differential equation will have two, and so on.

A second example

Let us yet again look at $\frac{{dy}}{{dx}} = \frac{{{x^3}}}{{{y^2}}}$

that we on the previous page found to have the solution ${y^3} = \frac{3}{4}{x^4} + C$

Let us find the solution going through the point (2, 3). We get ${3^3} = \frac{3}{4}{2^4} + C$

or 27=12+C, so C=15, and our particular solution will thus be ${y^3} = \frac{3}{4}{x^4} + 15$

More in general, to find a solution that goes through the point (x0, y0) we substitute that into our general solution to get $y_0^3 = \frac{3}{4}x_0^4 + C$

or $C = y_0^3 - \frac{3}{4}x_0^4$

This gives us the particular solution ${y^3} = \frac{3}{4}{x^4} - \frac{3}{4}x_0^4 + y_0^3$

or $y = {\left( {\frac{3}{4}({x^4} - x_0^4) + y_0^3} \right)^{1/3}}$

You can try this in this Geogebra file. Grab and move point A. Up a level : Differential Equations
Previous page : Separable differential equations - A few illustrating examples
Next page : Homogenous functions Last modified: Mar 30, 2019 @ 19:06