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Question: The solution of the differential equation $\frac{dy}{dx}+x(2x+y)=x^{2}(2x+y)^{3}-2$ is (C being an a...

The solution of the differential equation dydx+x(2x+y)=x2(2x+y)3−2\frac{dy}{dx}+x(2x+y)=x^{2}(2x+y)^{3}-2 is (C being an arbitrary constant)

Answer

e−x2(2x+y)2−xe−x2+∫e−x2dx=C\frac{e^{-x^2}}{(2x+y)^2} - x e^{-x^2} + \int e^{-x^2} dx = C

Explanation

Solution

Let the given differential equation be dydx+x(2x+y)=x2(2x+y)3−2\frac{dy}{dx}+x(2x+y)=x^{2}(2x+y)^{3}-2 Rearrange the terms: dydx+2+x(2x+y)=x2(2x+y)3\frac{dy}{dx} + 2 + x(2x+y) = x^2(2x+y)^3 Let v=2x+yv = 2x+y. Then dvdx=d(2x+y)dx=2+dydx\frac{dv}{dx} = \frac{d(2x+y)}{dx} = 2 + \frac{dy}{dx}. Substituting vv and dvdx\frac{dv}{dx} into the equation: (dvdx−2)+2+xv=x2v3\left(\frac{dv}{dx} - 2\right) + 2 + xv = x^2 v^3 dvdx+xv=x2v3\frac{dv}{dx} + xv = x^2 v^3 This is a Bernoulli differential equation of the form dvdx+P(x)v=Q(x)vn\frac{dv}{dx} + P(x)v = Q(x)v^n, with P(x)=xP(x) = x, Q(x)=x2Q(x) = x^2, and n=3n=3. To solve this, divide by v3v^3: v−3dvdx+xv−2=x2v^{-3} \frac{dv}{dx} + xv^{-2} = x^2 Let z=v1−n=v1−3=v−2z = v^{1-n} = v^{1-3} = v^{-2}. Then dzdx=−2v−3dvdx\frac{dz}{dx} = -2v^{-3} \frac{dv}{dx}. So, v−3dvdx=−12dzdxv^{-3} \frac{dv}{dx} = -\frac{1}{2} \frac{dz}{dx}. Substitute this into the equation: −12dzdx+xz=x2-\frac{1}{2} \frac{dz}{dx} + xz = x^2 Multiply by -2 to get the standard linear form: dzdx−2xz=−2x2\frac{dz}{dx} - 2xz = -2x^2 This is a first-order linear differential equation of the form dzdx+P′(x)z=Q′(x)\frac{dz}{dx} + P'(x)z = Q'(x), with P′(x)=−2xP'(x) = -2x and Q′(x)=−2x2Q'(x) = -2x^2. The integrating factor (I.F.) is e∫P′(x)dx=e∫−2xdx=e−x2e^{\int P'(x) dx} = e^{\int -2x dx} = e^{-x^2}. Multiply the linear equation by the integrating factor: e−x2dzdx−2xe−x2z=−2x2e−x2e^{-x^2} \frac{dz}{dx} - 2xe^{-x^2} z = -2x^2 e^{-x^2} The left side is the derivative of the product ze−x2z e^{-x^2}: ddx(ze−x2)=−2x2e−x2\frac{d}{dx}(z e^{-x^2}) = -2x^2 e^{-x^2} Integrate both sides with respect to xx: ∫ddx(ze−x2)dx=∫−2x2e−x2dx\int \frac{d}{dx}(z e^{-x^2}) dx = \int -2x^2 e^{-x^2} dx ze−x2=∫−2x2e−x2dxz e^{-x^2} = \int -2x^2 e^{-x^2} dx To evaluate the integral on the right side, we use integration by parts. Let I=∫−2x2e−x2dxI = \int -2x^2 e^{-x^2} dx. Let u=xu = x and dv=−2xe−x2dxdv = -2x e^{-x^2} dx. Then du=dxdu = dx. To find vv, integrate dvdv: v=∫−2xe−x2dxv = \int -2x e^{-x^2} dx. Let w=−x2w = -x^2, so dw=−2xdxdw = -2x dx. v=∫ewdw=ew=e−x2v = \int e^w dw = e^w = e^{-x^2}. Using the integration by parts formula ∫udv=uv−∫vdu\int u dv = uv - \int v du: I=x(e−x2)−∫e−x2dxI = x (e^{-x^2}) - \int e^{-x^2} dx So, the equation becomes: ze−x2=xe−x2−∫e−x2dx+Cz e^{-x^2} = x e^{-x^2} - \int e^{-x^2} dx + C where C is the arbitrary constant of integration. Substitute back z=v−2=(2x+y)−2z = v^{-2} = (2x+y)^{-2}: (2x+y)−2e−x2=xe−x2−∫e−x2dx+C(2x+y)^{-2} e^{-x^2} = x e^{-x^2} - \int e^{-x^2} dx + C Multiply by ex2e^{x^2}: (2x+y)−2=x+Cex2−ex2∫e−x2dx(2x+y)^{-2} = x + C e^{x^2} - e^{x^2} \int e^{-x^2} dx This does not seem to match standard forms or lead to a simple solution without the integral of e−x2e^{-x^2}.

Let's re-examine the integral ∫−2x2e−x2dx\int -2x^2 e^{-x^2} dx. Consider the derivative of xe−x2x e^{-x^2}: ddx(xe−x2)=1⋅e−x2+x⋅(−2xe−x2)=e−x2−2x2e−x2\frac{d}{dx}(x e^{-x^2}) = 1 \cdot e^{-x^2} + x \cdot (-2x e^{-x^2}) = e^{-x^2} - 2x^2 e^{-x^2}. So, −2x2e−x2=ddx(xe−x2)−e−x2-2x^2 e^{-x^2} = \frac{d}{dx}(x e^{-x^2}) - e^{-x^2}. Integrating both sides: ∫−2x2e−x2dx=∫(ddx(xe−x2)−e−x2)dx=xe−x2−∫e−x2dx\int -2x^2 e^{-x^2} dx = \int \left(\frac{d}{dx}(x e^{-x^2}) - e^{-x^2}\right) dx = x e^{-x^2} - \int e^{-x^2} dx. This is the same result.

Let's go back to ze−x2=xe−x2−∫e−x2dx+Cz e^{-x^2} = x e^{-x^2} - \int e^{-x^2} dx + C. Divide by e−x2e^{-x^2}: z=x−ex2∫e−x2dx+Cex2z = x - e^{x^2} \int e^{-x^2} dx + C e^{x^2}. Substitute z=(2x+y)−2z = (2x+y)^{-2}: (2x+y)−2=x−ex2∫e−x2dx+Cex2(2x+y)^{-2} = x - e^{x^2} \int e^{-x^2} dx + C e^{x^2}. 1(2x+y)2=x+ex2(C−∫e−x2dx)\frac{1}{(2x+y)^2} = x + e^{x^2} (C - \int e^{-x^2} dx).

Let's rewrite the solution as e−x2(2x+y)−2−xe−x2+∫e−x2dx=Ce^{-x^2} (2x+y)^{-2} - x e^{-x^2} + \int e^{-x^2} dx = C. This is a valid general solution. The integral ∫e−x2dx\int e^{-x^2} dx is a non-elementary function. So the solution can be written as: e−x2(2x+y)2−xe−x2+∫e−x2dx=C\frac{e^{-x^2}}{(2x+y)^2} - x e^{-x^2} + \int e^{-x^2} dx = C.