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Question: The general solution of the differential equation $\frac{dy}{dx} + x(x+y) = x(x+y)^3 -1$ is:...

The general solution of the differential equation dydx+x(x+y)=x(x+y)3−1\frac{dy}{dx} + x(x+y) = x(x+y)^3 -1 is:

A

ln∣(x+y+1)(x+y−1)(x+y)4∣=x2+Cln \left| \frac{(x+y+1)(x+y-1)}{(x+y)^4} \right| = x^2 + C

B

ln∣(x+y+1)(x+y−1)(x+y)2∣=x2+Cln \left| \frac{(x+y+1)(x+y-1)}{(x+y)^2} \right| = x^2 + C

C

2ln∣(x+y+1)(x+y−1)(x+y)2∣=x2+C2ln \left| \frac{(x+y+1)(x+y-1)}{(x+y)^2} \right| = x^2 + C

D

ln∣(x+y+1)(x+y−1)(x+y)2∣=x+Cln \left| \frac{(x+y+1)(x+y-1)}{(x+y)^2} \right| = x + C

Answer

ln | (x+y+1)(x+y-1) / (x+y)^2 | = x^2 + C

Explanation

Solution

The given differential equation is: dydx+x(x+y)=x(x+y)3−1\frac{dy}{dx} + x(x+y) = x(x+y)^3 -1 Rearrange the terms to group the constant with the derivative: dydx+1+x(x+y)=x(x+y)3\frac{dy}{dx} + 1 + x(x+y) = x(x+y)^3 Let v=x+yv = x+y. Differentiating vv with respect to xx: dvdx=ddx(x+y)=1+dydx\frac{dv}{dx} = \frac{d}{dx}(x+y) = 1 + \frac{dy}{dx} Substitute vv and dvdx\frac{dv}{dx} into the rearranged differential equation: dvdx+xv=xv3\frac{dv}{dx} + xv = xv^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, where P(x)=xP(x) = x, Q(x)=xQ(x) = x, and n=3n=3.

To solve a Bernoulli equation, divide by vnv^n (in this case, v3v^3): v−3dvdx+xv−2=xv^{-3}\frac{dv}{dx} + xv^{-2} = x Now, let z=v1−n=v1−3=v−2z = v^{1-n} = v^{1-3} = v^{-2}. Differentiate zz with respect to xx: dzdx=−2v−3dvdx\frac{dz}{dx} = -2v^{-3}\frac{dv}{dx} From this, we can express v−3dvdxv^{-3}\frac{dv}{dx} as −12dzdx-\frac{1}{2}\frac{dz}{dx}. Substitute this into the equation: −12dzdx+xz=x-\frac{1}{2}\frac{dz}{dx} + xz = x Multiply the entire equation by -2 to convert it into a standard first-order linear differential equation form: dzdx−2xz=−2x\frac{dz}{dx} - 2xz = -2x This is a linear differential equation of the form dzdx+P′(x)z=Q′(x)\frac{dz}{dx} + P'(x)z = Q'(x), where P′(x)=−2xP'(x) = -2x and Q′(x)=−2xQ'(x) = -2x.

The integrating factor (I.F.) is given by e∫P′(x)dxe^{\int P'(x) dx}: I.F.=e∫−2xdx=e−x2\text{I.F.} = e^{\int -2x dx} = e^{-x^2} Multiply the linear differential equation by the integrating factor: e−x2dzdx−2xe−x2z=−2xe−x2e^{-x^2}\frac{dz}{dx} - 2xe^{-x^2}z = -2xe^{-x^2} The left side of the equation is the derivative of the product z⋅I.F.z \cdot \text{I.F.}: ddx(ze−x2)=−2xe−x2\frac{d}{dx}(z e^{-x^2}) = -2xe^{-x^2} Now, integrate both sides with respect to xx: ∫ddx(ze−x2)dx=∫−2xe−x2dx\int \frac{d}{dx}(z e^{-x^2}) dx = \int -2xe^{-x^2} dx ze−x2=∫−2xe−x2dxz e^{-x^2} = \int -2xe^{-x^2} dx To evaluate the integral on the right side, let u=−x2u = -x^2. Then du=−2xdxdu = -2x dx. ∫eudu=eu+C1=e−x2+C1\int e^u du = e^u + C_1 = e^{-x^2} + C_1 So, the solution in terms of zz is: ze−x2=e−x2+C1z e^{-x^2} = e^{-x^2} + C_1 Divide by e−x2e^{-x^2}: z=1+C1ex2z = 1 + C_1 e^{x^2} Now, substitute back z=v−2z = v^{-2} and v=x+yv = x+y: (x+y)−2=1+C1ex2(x+y)^{-2} = 1 + C_1 e^{x^2} 1(x+y)2=1+C1ex2\frac{1}{(x+y)^2} = 1 + C_1 e^{x^2} Rearrange the equation to match the options provided: 1(x+y)2−1=C1ex2\frac{1}{(x+y)^2} - 1 = C_1 e^{x^2} 1−(x+y)2(x+y)2=C1ex2\frac{1 - (x+y)^2}{(x+y)^2} = C_1 e^{x^2} Using the difference of squares formula, 1−(x+y)2=(1−(x+y))(1+(x+y))1-(x+y)^2 = (1-(x+y))(1+(x+y)): (1−x−y)(1+x+y)(x+y)2=C1ex2\frac{(1 - x - y)(1 + x + y)}{(x+y)^2} = C_1 e^{x^2} Take the natural logarithm of both sides. Note that (1−x−y)=−(x+y−1)(1 - x - y) = -(x+y-1), and since we are taking absolute values, the negative sign outside doesn't affect the result. ln⁡∣(1−x−y)(1+x+y)(x+y)2∣=ln⁡∣C1ex2∣\ln \left| \frac{(1 - x - y)(1 + x + y)}{(x+y)^2} \right| = \ln |C_1 e^{x^2}| ln⁡∣(x+y−1)(x+y+1)(x+y)2∣=ln⁡∣C1∣+ln⁡∣ex2∣\ln \left| \frac{(x+y-1)(x+y+1)}{(x+y)^2} \right| = \ln |C_1| + \ln |e^{x^2}| ln⁡∣(x+y−1)(x+y+1)(x+y)2∣=x2+C\ln \left| \frac{(x+y-1)(x+y+1)}{(x+y)^2} \right| = x^2 + C where C=ln⁡∣C1∣C = \ln |C_1| is an arbitrary constant.

Comparing this result with the given options, it matches option (b).