# Why does a homogeneous ODE need initial condition for a particular solution while a non-homogeneous ODE doesn't?

For homogeneous ODE's with constant coefficients you need intial conditions such as y(x0)=y0 and y'(x0)=y1 or some boundary values to find a particular solution because you need to find the constants c1 and c2 in the general solution equation y=c1 $$e^{r1x} +$$ c2 $$e^{r2x}$$ where r1 and r2 are roots of the auxillary equation.

But for non-homogeneous ODE's you can find a particular solution without having to know any intial conditions or boundary values.

Why ?

• The particular solution is usually generated from an ansatz and is only required to adhere to the functional form of the underlying governing differential eqiuation. Note also, that for ODEs, Initial and Boundary conditions are only applied after you construct the full solution as the sum of the general and particular solutions. It is only for PDEs that you need homogeneous BCs, if you use Separation of Variables. – Sharat V Chandrasekhar Apr 29 at 20:31
• yes, we will have to apply initial value conditions after you construct the full solution as the sum of the general and particular solutions to get "the particular solution" adhering to those initial conditions – theenigma017 Apr 29 at 20:38
• Even for PDE's you dont need homogeneous BCs, even with Separation of Variables, if you can invoke a transformation to ensure homogensous BCs in the separated ODEs. – Sharat V Chandrasekhar Apr 29 at 20:41
• "The particular solution is usually generated from an ansatz and is only required to adhere to the functional form of the underlying governing differential eqiuation." Yes it is generated from an ansatz but what initial condition does the 'particular solution' adhere to ?, say when finding particular solution of non-homogeneous ODE's with the undetermined coefficient method – theenigma017 Apr 29 at 20:48
• See my comment below Prof. Israel's answer. – Sharat V Chandrasekhar Apr 29 at 20:56

No, you don't need initial conditions to find a particular solution of the homogeneous case. For example, $$e^{r_1 x}$$ is a particular solution.
• My textbook just says "one such function that solves ay" + by' +cy = 0 is $y=e^{rx}$ " and it's from there the general formula for a solution with the 2 constants are obtained so I'm not sure how .$y=e^{rx}$ is a particular solution. Also when finding particular solution of non-homogeneous ODE's with the undetermined coefficient method, what initial condition does the 'particular solution' adhere to ? I'm confused – theenigma017 Apr 29 at 20:45
• $e^{rx}$ is one of the solutions - you will have 2 values of $r$ corresponding to each of the roots of the characteristic solution. The superposition of these solutions with (as yet) undetermined coefficients is the general solution. to your homogeneous ODE. If your RHS were non-zero, say $f(x)$, then you would have to seek a solution $y_p(x)$ in a functional form related to $f(x)$, such that $y_p(x)$ satisfies your ODE. You then add this to your general solution and apply initial or boundary conditions to evaluate the coefficients. – Sharat V Chandrasekhar Apr 29 at 20:49
• Let's say that $f(x)=Qx^2$. What you do is set $y_p(x) = K_o + K_1 x + K_2 x^2$ and evaluate $K_0$, $K_1$ and $K_2$ by equating like terms (constant, linear and quadratic) after differentiating $y_p(x)$ twice and plugging it into the non-homogeneous ODE. For example, $cK_2= Q$ is readily apparent. – Sharat V Chandrasekhar Apr 29 at 20:53