y’=x²e^x + 2xe^x
To solve the given differential equation, we can use the method of integrating factors
To solve the given differential equation, we can use the method of integrating factors.
First, let’s write the given equation in standard form:
y’ – (x^2)e^x – 2xe^x = 0
Now, notice that the coefficient of y’ is 1, which is already in the desired form. We can focus on the terms involving y.
To find the integrating factor (IF), we multiply the entire equation by the IF, which is defined as the exponential of the integral of the coefficient of y with respect to x. In this case, the coefficient of y is 0, so the integral of 0 with respect to x is simply a constant C.
IF = e^(C) = e^0 = 1
Multiplying the original equation by the integrating factor, we get:
1(y’ – (x^2)e^x – 2xe^x) = 1(0)
Simplifying, we have:
y’ – (x^2)e^x – 2xe^x = 0
Now, let’s rearrange the equation:
y’ = (x^2)e^x + 2xe^x
Notice that the right side of the equation is the same as the original differential equation. This means that the integrating factor method does not apply here. We will instead solve this equation using the method of the particular solution and the complementary solution.
To find the particular solution, we need to guess a particular solution that meets the equation. Based on the form of the right side of the equation, we can guess that the particular solution will be of the form y_p = kj(x)e^x, where k is a constant to be determined and j(x) is a polynomial of order 2 since the highest power of x in the equation is x^2.
Plugging this guess into the equation, we get:
kj'(x)e^x + kj(x)e^x – (x^2)e^x – 2xe^x = 0
To simplify this equation, we can factor out the e^x:
e^x(kj'(x) + kj(x) – (x^2) – 2x) = 0
Since e^x is never zero, we can set the expression within the parentheses equal to zero:
kj'(x) + kj(x) – (x^2) – 2x = 0
Now, we need to determine the function j(x) that satisfies this equation.
Let’s expand kj'(x) using the product rule:
kj'(x) = (k(x^2)e^x)’ = k'(x^2)e^x + 2kxe^x
Substituting this back into the equation, we have:
k'(x^2)e^x + 2kxe^x + kj(x) – (x^2) – 2x = 0
Now, we can rearrange the terms:
k'(x^2)e^x + kj(x) + 2kxe^x – (x^2) – 2x = 0
The terms involving k(x) should cancel each other out. To achieve this, we can set:
k'(x^2)e^x + 2kxe^x = 0
Divide the equation by e^x:
k'(x^2) + 2kx = 0
This is a separable differential equation, and we can solve it by separating the variables and integrating:
∫k'(x^2) dx + 2∫kx dx = ∫0 dx
∫k'(x^2) dx = -2∫kx dx
Integrating both sides, we have:
∫k'(x^2) dx = -2∫kx dx
Integrating, we get:
k(x^2) = -kx^2 + C
Now, we solve for k:
0 = -kx^2 + C – k(x^2)
0 = -2kx^2 + C
Setting the term with x^2 equal to zero, we have:
-2kx^2 = 0
Since x^2 is never zero, we can conclude that k = 0.
Therefore, the particular solution is y_p = 0.
Moving on to the complementary solution, we solve the homogeneous equation:
y’ = (x^2)e^x + 2xe^x
The homogeneous equation has the same form as the original equation. To solve it, we can use the method of separation of variables.
Separating the variables, we have:
dy / (x^2)e^x + 2xe^x = dx
Integrating both sides:
∫ dy / (x^2)e^x + 2xe^x = ∫ dx
To integrate the left side, we can use substitution:
Let u = x^2, then du = 2x dx
∫(1/u)e^x du = ∫ dx
Integrating, we get:
ln|u|e^x = x + C
Substituting back u = x^2, we have:
ln|x^2|e^x = x + C
Since ln(a * b) = ln(a) + ln(b), we can rewrite the equation as:
ln|x^2e^x| = x + C
Now, removing the absolute value, we have the complementary solution:
ln(x^2e^x) = x + C
In exponential form, this becomes:
x^2e^x = Ke^x, where K = e^C
Rearranging the equation, we have:
x^2 = K
Therefore, the complementary solution is y_c = K, where K is a constant.
Combining the particular and complementary solutions, the general solution to the differential equation is:
y = y_c + y_p = K + 0 = K,
where K is an arbitrary constant.
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