# Why do we need two linearly independent solutions for 2nd order linear ODE

Let we have a second-order homogeneous linear ODE with two initial conditions.

$y''+ p(x)y'+q(x)y=0$

$y(x_0)=K_0$ and $y'(x_0)=K_1$

Why do we need two linearly independent solutions to satisfy the IVP.

If we have only one solution what would happen?

Could you please explain?

• If $cosx$ is a solution, then $kcosx$ is automatically also a solution,therefore its not providing any new information on the equation. – Avrham Aton Apr 2 '15 at 10:17
• true. I edited my question. thanks. – 104078 Apr 2 '15 at 10:19

## 5 Answers

second order linear differential equation needs two linearly independent solutions so that it has a solution for any initial condition, say, $y(0) = a, y'(0) = b$ for arbitrary $a, b.$ from a mechanical point of view the position and the velocity can be prescribed independently.

Having two linearly independent solutions gives us the genral solution,that is the general form of all the possible solutions for the equation, whereas only one gives you only part of the possible solutions.

Consider for example, the simple equation $y''=0$ it has obvious two solutions $y_1=C$ $y_2=x$,therfore the genral solution is $y=ax+c$ having only one solution does not give this general form.

The set of all solutions of such a differential equation forms a vector space under the canonical operations. The dimension of that vector space is $2$ and hence two linearly independent solutions will form a basis for it.

As you have written the initial value problem (IVP), there are 2 parameters characterizing the wanted solution, $K_1$ and $K_2$. Variation of these 2 parameters gives a 2 dimensional manifold of IVP and their solutions.

Linear ODE now have the property that their solutions form a linear or at least affine space, the first for homogeneous, the second for general inhomogeneous problems. As such, they can be described by giving the basis of the (underlying) vector space, and each such basis has 2 elements. For instance those for the initial conditions $(K_1,K_2)=(0,1)$ and $(K_1,K_2)=(1,0)$.

The set of all solutions of $$Lx=x^{(n)}+p_{n-1}(t)x^{(n-1)}+\cdots+p_0(t)x^{(0)}=0,$$ where $p_0,\ldots,p_{n-1}\in C(I)$, is an $n-$dimensional space $X$. If $\tau\in I$, and $\varphi_j$, $\,j=1,\ldots,n$, is the solution of the initial value problem $$Lx=0, \quad x^{(i-1)}(\tau)=\delta_{ij}, \,\,i=1,\ldots,n,$$ then $B=\{\varphi_1,\ldots,\varphi_n\}$ is basis of $X$. In fact, if $\psi$ is the solution of the initial value problem $$Lx=0, \quad x^{(i-1)}(\tau)=\xi_i, \,\,i=1,\ldots,n,$$ then $\psi=\xi_1\varphi_1+\cdots+\xi_n\varphi_n$.