In this page, lets explore the variation of parameters method for solving a second order differential equation with examples.

Variation of Parameters Formula

Formula: Let u and v be two linearly independent solutions of y”+py’+qy=0. Then the general solution of y”+py’+qy=R obtained by using the method of variations of parameters is given by

$y=u \displaystyle\int \dfrac{vR}{W(v,u)} dx – v \displaystyle\int \dfrac{uR}{W(v,u)} dx$ $+C_1u+C_2v$

where C₁ and C₂ are arbitrary constants and $W(v,u)$ is the Wronskian of $u$ and $v$.

Solved Problems

Question: Using the method of variation of parameters, solve

$y”+y=\mathrm{cosec} x$.

Solution:

Let u and v be two solutions of $y”+y=0$.

The auxiliary equation is given by

m²+1 = 0.

⇒ m² = -1.

⇒ m = ±i (complex roots)

So the solution of $y”+y=0$ is given as follows y = c₁ cosx +c₂ sinx where c₁ and c₂ are arbitrary constants.

So let us assume that

u= cosx, v=sinx.

Now,

⇒ W(v,u) = -(sin²x+cos²x) = -1.

So using the method of variation of parameters, the solution of the given equation is as follows

$y = u \displaystyle\int \dfrac{vR}{W(v,u)} dx – v \displaystyle\int \dfrac{uR}{W(v,u)} dx$ $+C_1u+C_2v$

⇒ y = $\cos x \displaystyle\int \dfrac{\sin x \cdot \text{cosec}x}{-1} dx$ $- \sin x \displaystyle\int \dfrac{\cos x \text{cosec} x}{-1} dx$ $+C_1\cos x+C_2\sin x$

⇒ y = $-\cos x \displaystyle\int dx$ $+ \sin x \displaystyle\int \dfrac{\cos x}{\sin x} dx$ $+C_1\cos x+C_2\sin x$

⇒ y = $-x\cos x + \sin x \ln|\sin x|+C_1\cos x+C_2\sin x$

So the general solution of $y”+y=\text{cosec} x$ by variation of parameters is y = $-x\cos x + \sin x \ln|\sin x|+C_1\cos x+C_2\sin x$ where C₁ and C₂ are arbitrary constants.