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)