Question

4. dy/dx=2y^2+3xy
x^2

y=\frac{c x^{3}}{1-c x^{2}}\]
y=cx31+cx2
y=−cx31−cx2
y=−cx31+cx2

Answer 1

ANSWER on Question #79414 – Math – Differential Equations

QUESTION

Solve the differential equation

$$\frac{dy}{dx} = \frac{2y^2 + 3xy}{x^2}$$

Possible answers:

a)

$$y = \frac{cx^3}{1 - cx^2}$$

b)

$$y = \frac{cx^3}{1 + cx^2}$$

c)

$$y = -\frac{cx^3}{1 - cx^2}$$

d)

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

SOLUTION

$$\frac{dy}{dx} = \frac{2y^2 + 3xy}{x^2} \rightarrow \frac{dy}{dx} = \frac{2y^2}{x^2} + \frac{3xy}{x^2} \rightarrow \boxed{\frac{dy}{dx} = 2 \cdot \left(\frac{y}{x}\right)^2 + 3 \cdot \frac{y}{x}}$$

We introduce the substitution

$$u = \frac{y}{x} \rightarrow y = ux \rightarrow \frac{dy}{dx} = \frac{du}{dx} \cdot x + u \cdot 1 \rightarrow \boxed{\frac{dy}{dx} = \frac{du}{dx} \cdot x + u}$$

Then,

$$\left\{ \begin{array}{l} \frac{dy}{dx} = 2 \cdot \left(\frac{y}{x}\right)^2 + 3 \cdot \frac{y}{x} \\ \quad u = \frac{y}{x} \\ \quad \text{at } \frac{du}{dx} \cdot x + u = 2u^2 + 3u \rightarrow \frac{du}{dx} \cdot x = 2u^2 + 3u - u \rightarrow \frac{du}{dx} \cdot x = 2u^2 + 2u \rightarrow \\ \frac{dy}{dx} = \frac{du}{dx} \cdot x + u \end{array} \right.$$ $$\left. \frac{du}{dx} \cdot x = 2u^2 + 2u \right| \times \left(\frac{2 \cdot dx}{x \cdot (2u^2 + 2u)}\right) \rightarrow \frac{2 \cdot du}{2u^2 + 2u} = \frac{2 \cdot dx}{x} \rightarrow \frac{2 \cdot ((u + 1) - u)du}{2u(u + 1)} = \frac{2 \cdot dx}{x} \rightarrow$$ $$\left(\frac{1}{u} - \frac{1}{u + 1}\right) du = \left(\frac{2}{x}\right) dx \rightarrow \int \left(\frac{1}{u} - \frac{1}{u + 1}\right) du = \int \left(\frac{2}{x}\right) dx \rightarrow \ln |u| - \ln |u + 1| = 2 \cdot \ln |x| + \ln |c| \rightarrow$$ $$\ln \left| \frac{u}{u + 1} \right| = \ln |cx^2| \rightarrow \frac{u}{u + 1} = cx^2 \Bigg| \times (u + 1) \rightarrow u = cx^2 \cdot (u + 1) \rightarrow u = cx^2 \cdot u + cx^2 \rightarrow$$ $$u - cx^2 \cdot u = cx^2 \rightarrow u(1 - cx^2) = cx^2 \rightarrow \boxed{u = \frac{cx^2}{1 - cx^2}}$$

We recall that we introduced a substitution

$$u = \frac{y}{x}$$

Then,

$$\left\{ \begin{array}{l} u = \frac{cx^2}{1 - cx^2} \rightarrow \frac{y}{x} = \frac{cx^2}{1 - cx^2} \Bigg| \times (x) \rightarrow y = \frac{cx^3}{1 - cx^2} \\ u = \frac{y}{x} \end{array} \right.$$

Conclusion,

$$\boxed{\frac{dy}{dx} = \frac{2y^2 + 3xy}{x^2} \rightarrow y = \frac{cx^3}{1 - cx^2} - \text{ANSWER } a)}$$

ANSWER:

$$a) \ y = \frac{cx^3}{1 - cx^2}$$

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