Solve the differential equation.
-$- x ^ { 3 } y ^ { \prime } + 2 x ^ { 2 } y = y ^ { 2 }$
A) $y = \frac { x ^ { 2 } } { \ln x - C }$
B) $y = \frac { x ^ { 2 } } { \ln x + C }$
C) $y = \frac { x } { \ln x - C }$
D) $y = \frac { x } { \ln x + C }$

Question

Quizplus · Accepted Answer

We first notice that this is a Bernoulli equation, which can be solved by substitution.

Let's divide both sides by y^2:

$$\frac{-x^3}{y}y' + \frac{2x^2}{y} = 1$$

Now, we substitute z = y^-1 and z' = -y^-2 y':

$$\frac{x^3}{z'}\left(-\frac{z'}{z^2}\right) + \frac{2x^2}{z'} = 1$$

Simplifying and multiplying both sides by z^2:

$$-x^3 z' + 2x^2 z = z^2$$

This is a separable differential equation, so let's put all the z terms on one side and all the x terms on the other:

$$\frac{z'}{z^2 - z} = -\frac{1}{x}$$

Now, we can integrate both sides. For the left side, we use partial fractions:

$$\frac{1}{z^2-z} = \frac{1}{z-1} - \frac{1}{z}$$

So we have:

$$\ln|z-1| - \ln|z| = -\ln|x| + C$$

Simplifying and using the fact that z = y^-1:

$$\ln\left|\frac{y}{y-1}\right| = \ln|x| + C$$

Taking the exponent of both sides:

$$\left|\frac{y}{y-1}\right| = e^{\ln|x|+C} = C|x|$$

Now, we use the fact that the absolute value of a quotient is the quotient of the absolute values:

$$\frac{y}{y-1} = \pm Cx$$

We solve for y:

$$y = \frac{Cx}{1 \pm Cx} = \frac{x}{\frac{1}{C}\pm x}$$

Comparing with the answer choices, we see that B matches with C = 1/C, which simplifies to C = 1. So the final answer is:

$$y = \frac{x^2}{\ln x + C}$$

Solve the Differential Equation −x3y′+2x2y=y2- x ^ { 3 } y ^ { \prime } + 2 x ^ { 2 } y = y ^ { 2 }−x3y′+2x2y=y2

Solve the Differential Equation $- x ^ { 3 } y ^ { \prime } + 2 x ^ { 2 } y = y ^ { 2 }$