Solved on Feb 26, 2024

Find the value of y0(2)y_0(2) where y0y_0 is the solution to the ODE xy+2y=2xxy' + 2y = 2x with y(1)=3y(1) = 3.

STEP 1

Assumptions
1. The given differential equation is xy+2y=2xxy' + 2y = 2x.
2. The initial condition is y(1)=3y(1) = 3.
3. We need to find the value of y0(2)y_0(2), where y0y_0 is the solution to the differential equation satisfying the initial condition.

STEP 2

To solve the differential equation, we first need to identify its type. The given equation is a first-order linear differential equation in the form xy+p(x)y=q(x)xy' + p(x)y = q(x), where p(x)=2/xp(x) = 2/x and q(x)=2q(x) = 2.

STEP 3

The integrating factor, μ(x)\mu(x), for a first-order linear differential equation is given by
μ(x)=ep(x)dx\mu(x) = e^{\int p(x)dx}

STEP 4

Calculate the integrating factor using the function p(x)=2/xp(x) = 2/x.
μ(x)=e2xdx\mu(x) = e^{\int \frac{2}{x}dx}

STEP 5

Integrate 2/x2/x to find the integrating factor.
μ(x)=e2lnx\mu(x) = e^{2\ln|x|}

STEP 6

Simplify the integrating factor using the properties of logarithms and exponentials.
μ(x)=elnx2\mu(x) = e^{\ln|x|^2}
μ(x)=x2\mu(x) = |x|^2
Since xx is in the domain where x>0x > 0 (because we are given the initial condition at x=1x = 1), we can drop the absolute value.
μ(x)=x2\mu(x) = x^2

STEP 7

Multiply the entire differential equation by the integrating factor x2x^2.
x2(xy+2y)=x2(2x)x^2(xy' + 2y) = x^2(2x)

STEP 8

Simplify the equation.
x3y+2x2y=2x3x^3y' + 2x^2y = 2x^3

STEP 9

Recognize that the left-hand side of the equation is the derivative of a product of functions.
ddx(x2y)=2x3\frac{d}{dx}(x^2y) = 2x^3

STEP 10

Integrate both sides of the equation with respect to xx.
ddx(x2y)dx=2x3dx\int \frac{d}{dx}(x^2y)dx = \int 2x^3dx

STEP 11

Perform the integration.
x2y=24x4+Cx^2y = \frac{2}{4}x^4 + C
x2y=12x4+Cx^2y = \frac{1}{2}x^4 + C

STEP 12

Solve for yy by dividing both sides by x2x^2.
y=12x2+Cx2y = \frac{1}{2}x^2 + \frac{C}{x^2}

STEP 13

Apply the initial condition y(1)=3y(1) = 3 to find the constant CC.
3=12(1)2+C(1)23 = \frac{1}{2}(1)^2 + \frac{C}{(1)^2}

STEP 14

Solve for CC.
3=12+C3 = \frac{1}{2} + C
C=312C = 3 - \frac{1}{2}
C=52C = \frac{5}{2}

STEP 15

Substitute the value of CC back into the equation for yy.
y=12x2+5/2x2y = \frac{1}{2}x^2 + \frac{5/2}{x^2}

STEP 16

Now that we have the general solution of the differential equation, we can find the value of y0(2)y_0(2) by plugging in x=2x = 2.
y0(2)=12(2)2+5/2(2)2y_0(2) = \frac{1}{2}(2)^2 + \frac{5/2}{(2)^2}

STEP 17

Calculate the value of y0(2)y_0(2).
y0(2)=12(4)+5/24y_0(2) = \frac{1}{2}(4) + \frac{5/2}{4}
y0(2)=2+58y_0(2) = 2 + \frac{5}{8}
y0(2)=168+58y_0(2) = \frac{16}{8} + \frac{5}{8}
y0(2)=218y_0(2) = \frac{21}{8}
The value of y0(2)y_0(2) is 218\frac{21}{8}.

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