Answer to Question #218276 in Differential Equations for Anuj

Question

Find the general solution and the second solution of,

x²y"-3xy'+4y=0,given y₁(x)=x²

Answer 1

The general solution of the differential equation will be in the form:

$$y=y_1(x)+y_2(x)$$

Since the solutions are linearly independent, we assume

$$y_2(x) = y_1(x)v(x) \implies y_2=vx^2\qquad \cdots (1)$$

Hence,

$$y_2'(x) = 2vx +v'x^2 \qquad\cdots (2)\\ y_2''(x) = 2v+4xv'+x^2v'' \quad \cdots (3)$$

Substitute (2) and (3) in (1):

$$x^2(2v+4xv'+x^2v'')-3x(2vx+v'x^2)+4vx^2=0$$

Simplifying the above:

$$2vx^2+4x^3v'+x^4v''-6vx^2-3v'x^3+4vx^2=0\\ x^4v''+x^3v'=0\\$$

Thus:

$$xv''+v'=0 \qquad \cdots (4)$$

Let:

$$w=v' \qquad\cdots (5)\\ w'=v'' \qquad\cdots (6)$$

Put (5) and (6) in (4)

$$xw'+w=0\\ xw'=-w\\ -\frac{w'}{w}=\frac{1}{x}$$

Integrate both sides

$$\int-\frac{w'}{w}dw=\int \frac{1}{x}dx\\ -\ln w =\ln x\\ \ln w^{-1} = \ln x\\ w^{-1} = x \implies w=\frac{1}{x}$$

Recall:

$$v'(x)=w \implies v=\int w dx = \int \frac{1}{x}dx = \ln x$$

Hence

$$v= \ln x \qquad \cdots (7)$$

Put (7) in (1), we get:

$$y_2 = x^2 \ln x$$

Which is the second solution.

The general solution is therefore:

$$y=C_1x^2+C_2x^2 \ln x$$