Solve the initial-value problem.x2y' + 2xy = ln(x), y(1) = 7
Question
Solve the initial-value problem.x2y' + 2xy = ln(x), y(1) = 7
Solution 1
This is a first order linear differential equation. We can solve it using an integrating factor.
Step 1: Rewrite the equation in standard form The standard form of a first order linear differential equation is y' + p(x)y = g(x). So, we divide the entire equation by x² to get it in standard form:
y' + (2/x)y = ln(x)/x²
Step 2: Find the integrating factor The integrating factor is e^(∫p(x)dx). Here, p(x) = 2/x. So, the integrating factor is e^(∫2/x dx) = e^(2ln|x|) = x².
Step 3: Multiply the equation by the integrating factor Multiplying every term of the equation by x², we get:
x²y' + 2xy = ln(x)
Step 4: The left side of this equation is the derivative of (x²y) with respect to x. So, we can write it as (x²y)' = ln(x).
Step 5: Integrate both sides with respect to x ∫(x²y)' dx = ∫ln(x) dx This gives us x²y = xln(x) - x + C.
Step 6: Solve for y y = (xln(x) - x + C) / x²
Step 7: Use the initial condition to find C Substitute x = 1 and y = 7 into the equation to solve for C:
7 = (1*ln(1) - 1 + C) / 1² This gives C = 8.
So, the solution to the differential equation is y = (xln(x) - x + 8) / x².
Solution 2
This is a first order linear differential equation. We can solve it using an integrating factor.
Step 1: Rewrite the equation in standard form The standard form of a first order linear differential equation is y' + p(x)y = g(x). So, we divide the entire equation by x^2 to get it in this form:
y' + (2/x)y = ln(x)/x^2
Step 2: Find the integrating factor The integrating factor is e^(∫p(x) dx). Here, p(x) = 2/x. So, the integrating factor is e^(∫2/x dx) = e^(2ln|x|) = x^2.
Step 3: Multiply the equation by the integrating factor Multiplying the equation by x^2, we get:
x^2y' + 2xy = ln(x)
Step 4: The left side of this equation is the derivative of (x^2)y with respect to x. So, we can write it as (d/dx)[x^2y] = ln(x).
Step 5: Integrate both sides with respect to x ∫(d/dx)[x^2y] dx = ∫ln(x) dx
This gives us x^2y = xln(x) - x + C.
Step 6: Solve for y y = (xln(x) - x + C) / x^2
Step 7: Use the initial condition to find C Substitute x = 1 and y = 7 into the equation to find C:
7 = (1*ln(1) - 1 + C) / 1^2 7 = -1 + C C = 8
So, the solution to the differential equation is y = (xln(x) - x + 8) / x^2.
Similar Questions
Y'=(y ln(y)+(yx^2))/-((x)+(2y^2)) ,y(3)=1 Find the solution of the following initial value problem.
To solve the given initial value problem: \[ y' - y = 7te^{2t}, \quad y(0) = 1 \] we can use the method of integrating factors. The differential equation is a first-order linear differential equation of the form: \[ y' - y = 7te^{2t} \] First, we identify the integrating factor \( \mu(t) \): \[ \mu(t) = e^{\int -1 \, dt} = e^{-t} \] Next, we multiply both sides of the differential equation by the integrating factor: \[ e^{-t} y' - e^{-t} y = 7te^{2t} e^{-t} \] \[ e^{-t} y' - e^{-t} y = 7t e^{t} \] The left-hand side of the equation is the derivative of \( e^{-t} y \): \[ \frac{d}{dt} (e^{-t} y) = 7t e^{t} \] Now, we integrate both sides with respect to \( t \): \[ \int \frac{d}{dt} (e^{-t} y) \, dt = \int 7t e^{t} \, dt \] The left-hand side integrates to: \[ e^{-t} y \] For the right-hand side, we use integration by parts. Let: \[ u = t \quad \text{and} \quad dv = 7e^{t} \, dt \] \[ du = dt \quad \text{and} \quad v = 7e^{t} \] Then: \[ \int 7t e^{t} \, dt = 7t e^{t} - \int 7e^{t} \, dt \] \[ = 7t e^{t} - 7e^{t} \] \[ = 7e^{t} (t - 1) \] So, we have: \[ e^{-t} y = 7e^{t} (t - 1) + C \] Multiplying both sides by \( e^{t} \): \[ y = 7e^{t} (t - 1) e^{t} + C e^{t} \] \[ y = 7e^{2t} (t - 1) + C e^{t} \] Now, we use the initial condition \( y(0) = 1 \): \[ 1 = 7e^{0} (0 - 1) + C e^{0} \] \[ 1 = -7 + C \] \[ C = 8 \] Therefore, the solution to the initial value problem is: \[ y(t) = 7e^{2t} (t - 1) + 8e^{t} \] So, the final solution is: \[ y(t) = 7te^{2t} - 7e^{2t} + 8e^{t} \]
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