3.0 Differential equation of first order and higher degrees

As we know that first order differential equations is of the form $f(x,y,p)$

where $p = \frac{{dy}}{{dx}}$. In some questions we will find out that the degree of $p$ is not always one and the differential equation of higher

degree can be written as $${p^n} + {F_1}(x,y){p^{n - 1}} + ... + {F_n}(x,y) = 0$$

The equations of these type are solved by one of the following methods:

If the equation is solvable for $p$, then the given expression can be resolved into $n$ linear factors and equating each of the linear factors equal to zero,

we get $n$ differential equations of first order and first degree.

The solutions for each of them is find out using the methods explained in the previous section.

Let these solutions be ${\phi _1}(x,y,{c_1})=0$, ${\phi _2}(x,y,{c_2})=0$,...,${\phi _n}(x,y,{c_n})=0$.

Hence, the general solution is given by $${\phi _1}(x,y,c),{\phi _2}(x,y,c),...,{\phi _n}(x,y,c) = 0$$

The differential equation $$y=px+f(p)$$ is known as Clairaut's equation. The solution of any equation of this form or converted to this form is

$$y=cx+f(c)$$ which is obtained by replacing $p$ i.e., $\frac{{dy}}{{dx}}$ by $c$ in the given differential equation.

Question 14. Solve $y = px + \frac{p}{{\sqrt {1 + {p^2}} }}$

Solution: Since the given differential equation is a clairaut's equation. Therefore, the solution is $$y = cx + \frac{c}{{\sqrt {1 + {c^2}} }}$$

Question 15. Solve ${y^2}\log y = pxy + {p^2}$

Solution: Let $\log y = t$ and on differentiating it, we get $\frac{1}{y}\frac{{dy}}{{dx}} = \frac{{dt}}{{dx}}$. If $\frac{{dt}}{{dx}} = p \Rightarrow \frac{p}{y} = p$.

Substituting these in the given equation,

we get $${y^2}t = y.pxy + {p^2}{y^2} $$$$ \Rightarrow t = px + {p^2} $$ which is the clairaut's form. Therefore, the solution is $$t = cx + {c^2} \Rightarrow \log y = cx + {c^2} $$