Question

Find the general and singular solutions of y=px+ap(1-p)​

Answer 1

Answer:

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Step-by-step explanation:

The given equation is y = px + (a/p), ......(1) Which is in Clairaut's form. So replacing p by c in (1) the solution is y = cx + (a/c) or c2x - yc + a = 0. ........(2) Now, c - discriminant relation of (2) is B2 - 4AC = 0; i.e., (-y)2 - 4xa = 0 or y2 = 4ax .......(3) Now, y2 = 4ax gives 2y(dy/dx) = 4a or p = 2a/y. Putting this value of p in (1), we get y = (2ax)/y + (y/2) or y2 = 4ax which is true by (3) satisfies (1) so y2 = 4ax is the required singular solution.

Answer 2

Answer:

The general solution is $p=C$.

The singular solution is $y=C(x+a-aC)$.

Step-by-step explanation:

Step 1 of 2

For general solution,

Consider the equation as follows:

$y=px+ap(1-p)$ . . . . . (1)

Write the equation (1) as follows:

$y=px+ap-ap^2$, where $p=\frac{dy}{dx}$

Differentiate both the sides with respect to $x$ as follows:

$\frac{dy}{dx}=\frac{d}{dx}(px) + \frac{d}{dx}(ap)-\frac{d}{dx}(ap^2)$

Simplify using chain rule.

$\frac{dy}{dx}=p\frac{d}{dx}(x) +x\frac{d}{dx}(p) + a\frac{d}{dx}(p)-a\frac{d}{dx}(p^2)$, where $a$ is constant.

$\frac{dy}{dx}=p +x\frac{dp}{dx} + a\frac{dp}{dx}-2ap\frac{dp}{dx}$

$p=p +(x+a-2ap)\frac{dp}{dx}$ (Since $p=\frac{dy}{dx}$)

Further,

$(x+a-2ap)\frac{dp}{dx}=0$

⇒ $\frac{dp}{dx}=0$

Integrate both the sides with respect $x$ as follows:

⇒ $p=C$, where $C$ is the integration constant.

Therefore, the general solution is $p=C$.

Step 2 of 2

For singular solution,

Now, substitute the value $C$ for $p$ in the equation (1) as follows:

⇒ $y=Cx+aC(1-C)$

Simplify as follows:

⇒ $y=C(x+a-aC)$

Therefore, the singular solution is $y=C(x+a-aC)$.

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