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Math  /  Calculus

Questionxy′+2xy+2y=xe−2x y^{\prime}+2 x y+2 y=x e^{-2}

Studdy Solution

STEP 1

1. The given equation is a first-order linear differential equation.
2. We will use an integrating factor to solve the differential equation.

STEP 2

1. Rewrite the equation in standard linear form.
2. Determine the integrating factor.
3. Multiply through by the integrating factor.
4. Integrate both sides.
5. Solve for y y .

STEP 3

Rewrite the differential equation in standard linear form, which is y′+P(x)y=Q(x) y' + P(x) y = Q(x) .
Given: xy′+2xy+2y=xe−2 x y^{\prime} + 2x y + 2y = x e^{-2}
Divide every term by x x to isolate y′ y' :
y′+2xxy+2xy=e−2 y^{\prime} + \frac{2x}{x} y + \frac{2}{x} y = e^{-2}
Simplify:
y′+2y+2xy=e−2 y^{\prime} + 2y + \frac{2}{x} y = e^{-2}

STEP 4

Identify P(x) P(x) from the equation y′+P(x)y=Q(x) y' + P(x) y = Q(x) . Here, P(x)=2+2x P(x) = 2 + \frac{2}{x} .
Calculate the integrating factor μ(x)=e∫P(x) dx \mu(x) = e^{\int P(x) \, dx} .
P(x)=2+2x P(x) = 2 + \frac{2}{x}
Integrate P(x) P(x) :
∫P(x) dx=∫(2+2x) dx=2x+2ln⁡∣x∣ \int P(x) \, dx = \int \left(2 + \frac{2}{x}\right) \, dx = 2x + 2\ln|x|
Thus, the integrating factor is:
μ(x)=e2x+2ln⁡∣x∣=e2x⋅∣x∣2 \mu(x) = e^{2x + 2\ln|x|} = e^{2x} \cdot |x|^2

STEP 5

Multiply the entire differential equation by the integrating factor μ(x)=e2x⋅∣x∣2 \mu(x) = e^{2x} \cdot |x|^2 :
e2x⋅∣x∣2⋅y′+e2x⋅∣x∣2⋅(2y+2xy)=e2x⋅∣x∣2⋅e−2 e^{2x} \cdot |x|^2 \cdot y^{\prime} + e^{2x} \cdot |x|^2 \cdot (2y + \frac{2}{x} y) = e^{2x} \cdot |x|^2 \cdot e^{-2}
Simplify:
e2x⋅∣x∣2⋅y′+2e2x⋅∣x∣2⋅y+2e2x⋅∣x∣⋅y=e2x⋅∣x∣2⋅e−2 e^{2x} \cdot |x|^2 \cdot y^{\prime} + 2e^{2x} \cdot |x|^2 \cdot y + 2e^{2x} \cdot |x| \cdot y = e^{2x} \cdot |x|^2 \cdot e^{-2}

STEP 6

Recognize the left side as the derivative of a product:
ddx(e2x⋅∣x∣2⋅y)=e2x⋅∣x∣2⋅e−2 \frac{d}{dx} \left( e^{2x} \cdot |x|^2 \cdot y \right) = e^{2x} \cdot |x|^2 \cdot e^{-2}
Integrate both sides with respect to x x :
∫ddx(e2x⋅∣x∣2⋅y) dx=∫e2x⋅∣x∣2⋅e−2 dx \int \frac{d}{dx} \left( e^{2x} \cdot |x|^2 \cdot y \right) \, dx = \int e^{2x} \cdot |x|^2 \cdot e^{-2} \, dx

STEP 7

Solve the integral:
e2x⋅∣x∣2⋅y=∫e2x⋅∣x∣2⋅e−2 dx+C e^{2x} \cdot |x|^2 \cdot y = \int e^{2x} \cdot |x|^2 \cdot e^{-2} \, dx + C
e2x⋅∣x∣2⋅y=∫∣x∣2 dx+C e^{2x} \cdot |x|^2 \cdot y = \int |x|^2 \, dx + C
e2x⋅∣x∣2⋅y=x33+C e^{2x} \cdot |x|^2 \cdot y = \frac{x^3}{3} + C

STEP 8

Solve for y y :
y=x33+Ce2x⋅∣x∣2 y = \frac{\frac{x^3}{3} + C}{e^{2x} \cdot |x|^2}
y=x3e2x+Ce2x⋅∣x∣2 y = \frac{x}{3e^{2x}} + \frac{C}{e^{2x} \cdot |x|^2}
The solution for y y is:
y=x3e2x+Ce2x⋅∣x∣2 y = \frac{x}{3e^{2x}} + \frac{C}{e^{2x} \cdot |x|^2}

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