xy'-y=2xlnx
Solve the differential equations
y = cx + x ln^2(x)
To solve the given differential equation xy' - y = 2xln(x), we can use the method of integrating factors. The general form of a first-order linear differential equation is y' + P(x)y = Q(x), where P(x) and Q(x) are functions of x.
1. Rewrite the equation in the standard form: y' - (1/x)y = 2ln(x).
Comparing this with the standard form y' + P(x)y = Q(x), we have P(x) = -1/x and Q(x) = 2ln(x).
2. Calculate the integrating factor (IF):
The integrating factor is given by the equation IF = e^(∫P(x)dx).
Integrating P(x) = -1/x with respect to x, we have ∫P(x)dx = -ln(x).
Therefore, the integrating factor IF = e^(-ln(x)) = 1/x.
3. Multiply the entire differential equation by the integrating factor (IF):
(1/x)(xy') - (1/x)(y) = 2ln(x)(1/x).
Simplifying, we get y' - (1/x)y = 2ln(x)/x.
4. Rewrite the left side as the derivative of the product of y and IF:
(y/x)' = 2ln(x)/x.
5. Integrate both sides with respect to x:
∫(y/x)' dx = ∫2ln(x)/x dx.
The left side can be simplified using the property of logarithms:
∫(y/x)' dx = ∫1/x d(y/x) = ln|x| + C1,
where C1 is the constant of integration.
The right side can be integrated using the substitution method:
Let u = ln(x), then du = 1/x dx.
Substituting u and du into the integral, we get:
∫2ln(x)/x dx = 2∫u du = u^2 + C2,
where C2 is another constant of integration.
6. Combine the integration results:
ln|x| + C1 = u^2 + C2,
Rewriting u^2 = (ln(x))^2 and combining the constants of integration, we have:
ln|x| = (ln(x))^2 + C3,
where C3 = C2 - C1 is a new constant.
7. Solve for y:
Taking the exponential of both sides, we get:
|x| = e^((ln(x))^2 + C3).
Evaluating the right side of the equation:
|x| = e^(ln(x)^2) * e^(C3),
|x| = (e^(ln(x)))^2 * e^(C3),
|x| = x^2 * e^(C3).
Since |x| = x for positive values of x and |x| = -x for negative values of x, we consider the two cases separately:
For x > 0:
x = x^2 * e^(C3),
1 = x * e^(C3),
e^(C3) = 1/x,
C3 = -ln(x).
For x < 0:
-x = x^2 * e^(C3),
-1 = x * e^(C3),
e^(C3) = -1/x,
C3 = ln(-1) - ln(x) = ln(1/x).
Combining the two cases, we have:
C3 = ln(1/x).
Therefore, the general solution to the given differential equation is:
y = x^2 * e^(ln(1/x)).
Finally, simplifying the expression:
y = x^2/x = x.