- #1
Ted123
- 428
- 0
One solution of the differential equation
x^2(x^2+1)y^{\prime\prime} - 2x^3 y^{\prime} + 2(x^2-1) y = 0
can be obtained in the form y_1 = x^n. Use this solution to find another, and in this way find the general solution.
The DE can be written as:
\displaystyle y^{\prime\prime} - \frac{2x}{x^2+1} y^{\prime} + \frac{2(x^2-1)}{x^2(x^2+1)}y = 0.
\displaystyle - \int \frac{2x}{x^2+1} = -\ln (x^2+1)
Therefore the Wronskian \displaystyle W(x) = Ce^{\ln (x^2+1)} = C(x^2+1).
By inspection y_1 = x^2 - how can you see this straight away?
To find y_2 use the formula below with W for C=1 - can you always just take C=1?
\displaystyle y_2 = y_1 \int \frac{W}{y_1^2}\;dx
x^2(x^2+1)y^{\prime\prime} - 2x^3 y^{\prime} + 2(x^2-1) y = 0
can be obtained in the form y_1 = x^n. Use this solution to find another, and in this way find the general solution.
The DE can be written as:
\displaystyle y^{\prime\prime} - \frac{2x}{x^2+1} y^{\prime} + \frac{2(x^2-1)}{x^2(x^2+1)}y = 0.
\displaystyle - \int \frac{2x}{x^2+1} = -\ln (x^2+1)
Therefore the Wronskian \displaystyle W(x) = Ce^{\ln (x^2+1)} = C(x^2+1).
By inspection y_1 = x^2 - how can you see this straight away?
To find y_2 use the formula below with W for C=1 - can you always just take C=1?
\displaystyle y_2 = y_1 \int \frac{W}{y_1^2}\;dx
