Question

solve the given differential equation. (9x + 1)y^{2}\frac{dy}{dx}+4x^{2}+3y^{3}=0

Explanation:

Step1: Rearrange the differential equation

We rewrite the given equation \((9x + 1)y^{2}\frac{dy}{dx}+4x^{2}+3y^{3}=0\) as \((9x + 1)y^{2}\frac{dy}{dx}=-4x^{2}-3y^{3}\), and then \(\frac{dy}{dx}=\frac{-4x^{2}-3y^{3}}{(9x + 1)y^{2}}\). This is a non - linear first - order differential equation. We try to separate variables. First, rewrite it as \((9x + 1)y^{2}dy=-(4x^{2}+3y^{3})dx\), and further as \((9x + 1)y^{2}dy+ (4x^{2}+3y^{3})dx = 0\). Let's try to find an integrating factor or use substitution. Let \(u = y^{3}\), then \(\frac{du}{dy}=3y^{2}\) and \(dy=\frac{du}{3y^{2}}\).
Substituting \(u = y^{3}\) into the differential equation:
The original equation \((9x + 1)y^{2}\frac{dy}{dx}+4x^{2}+3y^{3}=0\) becomes \(\frac{1}{3}(9x + 1)\frac{du}{dx}+4x^{2}+u = 0\).

Step2: Rewrite in standard linear form

We rewrite \(\frac{1}{3}(9x + 1)\frac{du}{dx}+u=-4x^{2}\) as \(\frac{du}{dx}+\frac{3}{9x + 1}u=\frac{-12x^{2}}{9x + 1}\).
The integrating factor \(I.F.=e^{\int\frac{3}{9x + 1}dx}\). Let \(t = 9x+1\), then \(dx=\frac{1}{9}dt\) and \(\int\frac{3}{9x + 1}dx=\int\frac{3}{t}\times\frac{1}{9}dt=\frac{1}{3}\ln|t|=\frac{1}{3}\ln|9x + 1|\). So, \(I.F.=(9x + 1)^{\frac{1}{3}}\).

Step3: Multiply by integrating factor

Multiply the linear differential equation \(\frac{du}{dx}+\frac{3}{9x + 1}u=\frac{-12x^{2}}{9x + 1}\) by \((9x + 1)^{\frac{1}{3}}\):
\((9x + 1)^{\frac{1}{3}}\frac{du}{dx}+(9x + 1)^{-\frac{2}{3}}\times3u=\frac{-12x^{2}}{(9x + 1)^{\frac{2}{3}}}\).
The left - hand side is the derivative of \((9x + 1)^{\frac{1}{3}}u\) with respect to \(x\) by the product rule. So, \(\frac{d}{dx}[(9x + 1)^{\frac{1}{3}}u]=\frac{-12x^{2}}{(9x + 1)^{\frac{2}{3}}}\).

Step4: Integrate both sides

Integrate \(\int\frac{d}{dx}[(9x + 1)^{\frac{1}{3}}u]dx=\int\frac{-12x^{2}}{(9x + 1)^{\frac{2}{3}}}dx\).
Let \(9x+1 = z\), then \(x=\frac{z - 1}{9}\) and \(dx=\frac{1}{9}dz\).
\(x^{2}=\frac{(z - 1)^{2}}{81}\), and \(\int\frac{-12x^{2}}{(9x + 1)^{\frac{2}{3}}}dx=\int\frac{-12}{81}\times\frac{(z - 1)^{2}}{z^{\frac{2}{3}}}\times\frac{1}{9}dz=\frac{-4}{243}\int\frac{z^{2}-2z + 1}{z^{\frac{2}{3}}}dz=\frac{-4}{243}\int(z^{\frac{4}{3}}-2z^{\frac{1}{3}}+z^{-\frac{2}{3}})dz\).
\(\int(z^{\frac{4}{3}}-2z^{\frac{1}{3}}+z^{-\frac{2}{3}})dz=\frac{3}{7}z^{\frac{7}{3}}-\frac{3}{2}z^{\frac{4}{3}}+3z^{\frac{1}{3}}+C\).
After substituting back \(z = 9x + 1\) and solving for \(u\) (since \(u = y^{3}\)), we get the solution.
Another way:
We rewrite the original equation \((9x + 1)y^{2}\frac{dy}{dx}+4x^{2}+3y^{3}=0\) as \((9x + 1)y^{2}\frac{dy}{dx}=- (4x^{2}+3y^{3})\).
Let \(v=\frac{y}{x}\), then \(y = vx\) and \(\frac{dy}{dx}=v+x\frac{dv}{dx}\).
Substitute \(y = vx\) into the original equation:
\((9x + 1)(vx)^{2}(v + x\frac{dv}{dx})=-4x^{2}-3(vx)^{3}\).
\((9x + 1)v^{2}x^{2}(v + x\frac{dv}{dx})=-4x^{2}-3v^{3}x^{3}\).
\((9x + 1)v^{2}(v + x\frac{dv}{dx})=- \frac{4}{x}-3v^{3}\).
\((9x + 1)v^{3}+(9x + 1)v^{2}x\frac{dv}{dx}=- \frac{4}{x}-3v^{3}\).
\((9x + 1)v^{2}x\frac{dv}{dx}=- \frac{4}{x}-3v^{3}-(9x + 1)v^{3}\).
\((9x + 1)v^{2}x\frac{dv}{dx}=- \frac{4}{x}-(9x + 4)v^{3}\).
\(\frac{(9x + 1)v^{2}}{(9x + 4)v^{3}+\frac{4}{x}}dv=-\frac{1}{x}dx\).
Integrating both sides:
\(\int\frac{(9x + 1)v^{2}}{(9x + 4)v^{3}+\frac{4}{x}}dv=-\int\frac{1}{x}dx\).
Let \(w=(9x + 4)v^{3}+\frac{4}{x}\), then \(dw=(27xv^{3}+(9x + 4)\times3v^{2}x\frac{dv}{dx}-\frac{4}{x^{2}})dx\).
After integrating and substituting back \(v=\frac{y}{x}\), we get the general solution.

The general solution of the differential equation \((9x + 1)y^{2}\frac{dy}{dx}+4x^{2}+3y^{3}=0\) is \(y^{3}(9x + 1)^{\frac{1}{3}}+\frac{4}{3}(9x + 1)^{\frac{1}{3}}\int\frac{x^{2}}{(9x + 1)^{\frac{2}{3}}}dx=C\)

Answer:

\(y^{3}(9x + 1)^{\frac{1}{3}}+\frac{4}{3}(9x + 1)^{\frac{1}{3}}\int\frac{x^{2}}{(9x + 1)^{\frac{2}{3}}}dx=C\)