Solve the following 2nd order differential equation explicitly by expanding up to order x5,
y''(x) = (1 + x^2)*y(x)
pls show working y =f(x)
To solve the given 2nd order differential equation explicitly by expanding up to order x^5, we can use the power series method. Let's assume that the solution can be written as a power series in terms of x:
y(x) = a_0 + a_1*x + a_2*x^2 + a_3*x^3 + a_4*x^4 + a_5*x^5 + ...
Now, let's differentiate y(x) twice and substitute it into the differential equation:
y''(x) = 2*a_2 + 6*a_3*x + 12*a_4*x^2 + 20*a_5*x^3 + ...
Substituting into the differential equation, we have:
2*a_2 + 6*a_3*x + 12*a_4*x^2 + 20*a_5*x^3 + ... = (1 + x^2)*(a_0 + a_1*x + a_2*x^2 + a_3*x^3 + a_4*x^4 + a_5*x^5 + ...)
Expanding the right-hand side of the equation and collecting like terms, we get:
2*a_2 + 6*a_3*x + 12*a_4*x^2 + 20*a_5*x^3 + ... = a_0 + a_1*x + (a_2 + a_0)*x^2 + (a_3 + a_1)*x^3 + (a_4 + a_2)*x^4 + (a_5 + a_3)*x^5 + ...
Now, we set the coefficients of each power of x on both sides of the equation equal to each other. For example, comparing the coefficients of x^0, we have:
2*a_2 = a_0
Comparing the coefficients of x^1, we have:
6*a_3 = a_1
Comparing the coefficients of x^2, we have:
12*a_4 = a_2 + a_0
Comparing the coefficients of x^3, we have:
20*a_5 = a_3 + a_1
Continuing this pattern, we can write down a system of equations for each coefficient. Using the given initial conditions or boundary conditions, we can then solve the system of equations to find the values of a_0, a_1, a_2, a_3, a_4, and a_5.
Since you want the solution up to order x^5, we can solve for a_0 to a_5 to obtain the explicit form of the solution y(x) = f(x).