Implicit Differential Equations
Solved Problems
Example 1.
Find the general solution of the equation \[9{\left( {y'} \right)^2} - 4x = 0.\]
Solution.
This equation is of type \(x = f\left( {y'} \right)\) (Case \(3\text{).}\) We introduce the parameter \(p = y'\) and write the equation in the form:
\[x = \frac{9}{4}{p^2}.\]
By taking differentials of both sides, we obtain:
\[dx = \frac{9}{4} \cdot 2pdp = \frac{9}{2}pdp.\]
Since \(dy = pdx,\) the last expression can be presented as
\[\frac{{dy}}{p} = \frac{9}{2}pdp,\;\; \Rightarrow dy = \frac{9}{2}{p^2}dp.\]
By integrating we find the dependence of the variable \(y\) on the parameter \(p:\)
\[y = \int {\frac{9}{2}{p^2}dp} = \frac{9}{2}\int {{p^2}dp} = \frac{9}{2} \cdot \frac{{{p^3}}}{3} + C = \frac{3}{2}{p^3} + C,\]
where \(C\) is a constant.
Thus, we get the general solution of the equation in parametric form:
\[\left\{ \begin{array}{l}
y = \frac{3}{2}{p^3} + C\\
x = \frac{9}{4}{p^2}
\end{array} \right..\]
We can eliminate the parameter \(p\) from this system. It follows from the second equation that
\[{p^2} = \frac{4}{9}x,\;\; \Rightarrow p = \pm \frac{2}{3}{x^{\frac{1}{2}}}.\]
Substituting this in the first equation, we obtain the general solution as the explicit function \(y = f\left( x \right):\)
\[y = \frac{3}{2}{\left( { \pm \frac{2}{3}{x^{\frac{1}{2}}}} \right)^3} + C
= \pm \frac{3}{2} \cdot \frac{8}{{27}}{x^{\frac{3}{2}}} + C
= \pm \frac{4}{9}{x^{\frac{3}{2}}} + C.\]
Example 2.
Find the general solution of the differential equation \[y = \ln \left[ {25 + {{\left( {y'} \right)}^2}} \right].\]
Solution.
This differential equation is related to Case \(1\) because it contains only the variable \(y\) and its derivative \(y'.\) Using the parameter \(p\) we rewrite this equation in the following way:
\[y = \ln \left( {25 + {p^2}} \right).\]
Take the differentials of both sides:
\[dy = \frac{{2pdp}}{{25 + {p^2}}}.\]
As \(dy = pdx,\) we get
\[pdx = \frac{{2pdp}}{{25 + {p^2}}},\;\; \Rightarrow dx = \frac{{2dp}}{{25 + {p^2}}}.\]
Now we can integrate the last expression to obtain \(x\) as function of \(p.\)
\[x = \int {\frac{{2dp}}{{25 + {p^2}}}}
= 2\int {\frac{{dp}}{{25 + {p^2}}}}
= 2 \cdot \frac{1}{5}\arctan \frac{p}{5} + C
= \frac{2}{5}\arctan \frac{p}{5} + C.\]
So we have the following parametric representation of the solution of the differential equation:
\[\left\{ \begin{array}{l}
x = \frac{2}{5}\arctan \frac{p}{5} + C\\
y = \ln \left( {25 + {p^2}} \right)
\end{array} \right.,\]
where \(C\) is an arbitrary constant.
Example 3.
Solve the differential equation \[2y = 2{x^2} + 4xy' + \left( {y'} \right)^2.\]
Solution.
This equation relates to Case \(2.\) Let \(y' = p,\) so we can rewrite the equation as
\[2y = 2{x^2} + 4xp + {p^2}.\]
Find the differentials of both sides taking into account that \(dy = pdx.\) This yields:
\[2dy = 4xdx + 4pdx + 4xdp + 2pdp,\;\;\Rightarrow
dy = 2xdx + 2pdx + 2xdp + pdp,\;\; \Rightarrow
\underline {pdx} = 2xdx + \underline {2pdx} + 2xdp + pdp,\;\; \Rightarrow
0 = 2xdx + pdx + 2xdp + pdp,\;\; \Rightarrow
\left( {2x + p} \right)dx + \left( {2x + p} \right)dp = 0,\;\; \Rightarrow
\left( {2x + p} \right)\left( {dx + dp} \right) = 0.\]
We have two solutions that satisfy the last equation:
\[\left. 1 \right)\;\;2x + p = 0.\]
Hence,
\[2x + y' = 0,\;\; \Rightarrow y' = - 2x,\;\; \Rightarrow dy = - 2xdx.\]
By integrating this simple equation, we obtain:
\[{y_1} = - {x^2} + C,\]
where \(C\) is a constant. To determine the value of \(C,\) we substitute this answer in the original differential equation:
\[{y_1} = - {x^2} + C,\;\; \Rightarrow
{y_1}^\prime = - 2x,\;\; \Rightarrow
2\left( { - {x^2} + C} \right) = 2{x^2} + 4x \cdot \left( { - 2x} \right) + {\left( { - 2x} \right)^2},\;\; \Rightarrow
- 2{x^2} + 2C = 2{x^2} - 8{x^2} + 4{x^2},\;\; \Rightarrow
2C = 0,\;\; \Rightarrow C = 0.\]
We see that the constant \(C\) must be equal to zero to satisfy the equation. Thus, the first solution is
\[{y_1} = - {x^2}.\]
Now we consider the second solution, which is defined by the differential equation
\[\left. 2 \right)\;\;dx + dp = 0.\]
Then
\[\int {dx} = - \int {dp} ,\;\; \Rightarrow x = - p + C.\]
At the beginning of the solution we have written the differential equation in the form
\[2y = 2{x^2} + 4xp + {p^2}.\]
We can substitute the known expression for \(x\) (as a function of the parameter \(p\)) to find the dependence of \(y\) on \(p:\)
\[2y = 2{\left( { - p + C} \right)^2} + 4\left( { - p + C} \right)p + {p^2},\;\; \Rightarrow
2y = 2\left( {{p^2} - 2pC + {C^2}} \right) - 4{p^2} + 4pC + {p^2},\;\;\Rightarrow
2y = 2{p^2} - \cancel{4pC} + 2{C^2} - 3{p^2} + \cancel{4pC},\;\; \Rightarrow
2y = 2{C^2} - {p^2},\;\; \Rightarrow
y = {C^2} - \frac{{{p^2}}}{2}.\]
Thus, the second solution is given parametrically by the following system:
\[\left\{ \begin{array}{l}
x = - p + C\\
y = {C^2} - \frac{{{p^2}}}{2}
\end{array} \right.,\]
where \(C\) is a constant. Eliminating the parameter \(p,\) we can write the explicit solution:
\[p = C - x,\;\; \Rightarrow
y = {C^2} - \frac{{{{\left( {C - x} \right)}^2}}}{2}
= {C^2} - \frac{{{{\left( {x - C} \right)}^2}}}{2}.\]
The final answer is given by
\[y = {C^2} - \frac{{{{\left( {x - C} \right)}^2}}}{2},\;\; y = - {x^2}.\]
