To solve the differential equation x²y” + 2y’ = e^x, we start by rewriting the equation in a more standard form:
y” + rac{2}{x^{2}} y’ = rac{e^x}{x^{2}}.
This is a second-order linear non-homogeneous differential equation where the homogeneous part is:
x²y” + 2y’ = 0.
To solve the homogeneous part, we can make a substitution. Assume a solution of the form y = x^{m}. Then:
y’ = mx^{m-1} and y” = m(m-1)x^{m-2}.
Substituting these expressions into the homogeneous equation yields:
x²(m(m-1)x^{m-2}) + 2(mx^{m-1}) = 0, which simplifies to:
m(m-1) + 2m = 0.
This gives us the characteristic equation:
m² + m = 0, which factors into:
m(m + 1) = 0.
Thus, the roots are m = 0 and m = -1. The general solution for the homogeneous equation is:
y_h = C_1 + rac{C_2}{x}, where C_1 and C_2 are constants.
Next, we need to find a particular solution y_p for the non-homogeneous equation using the method of undetermined coefficients. Since the right-hand side is e^x, we can try a solution of the form:
y_p = Ae^x, where A is a constant to be determined.
Calculating the derivatives gives:
y_p’ = Ae^x and y_p” = Ae^x.
Now substituting these into the non-homogeneous equation:
x²(Ae^x) + 2(Ae^x) = e^x.
This simplifies to:
(Ax² + 2A)e^x = e^x.
Setting the coefficients equal, we compare:
Ax² + 2A = 1.
We determine that for the equation to hold for all x, A must equal 0 for the x² term and 2A = 1. Thereby:
A = rac{1}{2}.
Thus, a particular solution is:
y_p = rac{1}{2}e^x.
Finally, the general solution of the differential equation is the sum of the homogeneous and particular solutions:
y = C_1 + rac{C_2}{x} + rac{1}{2}e^x.