2.20.3.3.1 Case when \(n=2a\) (section 3-3 a(i))

2.20.3.3.1.1 Example \(xy^{\prime }=cx^{4}+2y-by^{2}\) Given

\[ xy^{\prime }=cx^{n}+ay-by^{2}\]

Where \(n=2a\) as in this example. In this case let \(y=x^{a}u\). Then \(y^{\prime }=ax^{a-1}u+x^{a}u^{\prime }\) and the ode becomes

\begin{align} x\left ( ax^{a-1}u+x^{a}u^{\prime }\right ) & =cx^{2a}+a\left ( x^{a}u\right ) -b\left ( x^{2a}u^{2}\right ) \nonumber \\ ax^{a}u+x^{a+1}u^{\prime } & =cx^{2a}+a\left ( x^{a}u\right ) -b\left ( x^{2a}u^{2}\right ) \nonumber \\ x^{a+1}u^{\prime } & =cx^{2a}-ax^{a}u+ax^{a}u-b\left ( x^{2a}u^{2}\right ) \nonumber \\ x^{a+1}u^{\prime } & =cx^{2a}-bx^{2a}u^{2}\nonumber \\ u^{\prime } & =cx^{2a-a-1}-bx^{2a-a-1}u^{2}\nonumber \\ u^{\prime } & =cx^{a-1}-bx^{a-1}u^{2}\nonumber \\ & =x^{a-1}\left ( c-bu^{2}\right ) \tag {1}\end{align}

We see that (1) is separable. Solving (1) gives

\[ u=\frac {1}{b}\tanh \left ( \frac {\sqrt {cb}\left ( C_{1}a+x^{a}\right ) }{a}\right ) \sqrt {cb}\]

But \(y=x^{a}u\), hence \(u=yx^{-a}\). Therefore

\begin{align*} yx^{-a} & =\frac {1}{b}\tanh \left ( \frac {\sqrt {cb}\left ( C_{1}a+x^{a}\right ) }{a}\right ) \sqrt {cb}\\ y & =\sqrt {cb}\frac {x^{a}}{b}\tanh \left ( \frac {\sqrt {cb}\left ( C_{1}a+x^{a}\right ) }{a}\right ) \end{align*}

Is the final solution for \(xy=cx^{n}+ay-by^{2}\) when \(n=2a\). The above can be simplified more if we know the specific numerical values of \(b,c.\) Note that in method, we have to know the values of \(a,n\,\) always in order to decide.