Solve and find the general solution of y
(3x+1)y"-(12x+5y)y'+6y(2x+1)=28⁴√(3x+1)³.•e²x
Help me out please ......
You sure you didn't mean
(3x+1)y"-(12x+5)y'+6y(2x+1)=28*(3x+1)^(3/4)*e^(2x)
because e^2 * x is just very incongruous, and that 12x+5y makes it nonlinear.
Check out integrating factors.
I saw e^2 * x but I suspect it was a typo from them
But can you show me how to do it
Having e^(2x)
I will follow your step all through
To solve the given differential equation, we need to follow these steps:
1. Identify the type of the equation: The given equation is a second-order linear homogeneous ordinary differential equation since it involves the second derivative of the dependent variable 'y' and the coefficients of y, y', and y" are linear functions of x.
2. Rewrite the equation in standard form: We begin by dividing the entire equation by (3x+1) to eliminate the √(3x+1) term:
y" - [(12x + 5)/(3x + 1)]y' + 2[(2x + 1)/(3x + 1)]y = 28√(3x+1)³ • e^(2x) / (3x+1)
3. Find the general solution of the homogeneous equation (without the right-hand side): We ignore the right-hand side for now and consider the homogeneous equation:
y" - [(12x + 5)/(3x + 1)]y' + 2[(2x + 1)/(3x + 1)]y = 0
The roots of the characteristic equation (obtained by assuming y to be in the form of e^(rx)) will give us the solution:
r² - [(12x + 5)/(3x + 1)]r + 2[(2x + 1)/(3x + 1)] = 0
Solve this quadratic equation to find the values of r. Substituting these values back into y, we get the general solution of the homogeneous equation.
4. Find a particular solution for the non-homogeneous equation: Next, we consider the non-homogeneous equation:
y_p = A(x)√(3x+1)³ • e^(2x) / (3x+1)
To determine A(x), we substitute this particular solution back into the original equation and solve for A(x) by comparing coefficients. This involves some algebraic manipulation.
5. Obtain the general solution of the entire equation: The general solution of the given differential equation is the sum of the general solution of the homogeneous equation and the particular solution.