Question

10. Reduce the differential equation y = 2 px + y²p to Clairaut's form by putting
y? = v and hence find its general and singular solutions.
Solve and test​

Answer 1

Answer:

Let's put X=x^2;\quad Y=y^2X=x2;Y=y2 then p=\frac{x}{y}\frac{dY}{dX}p=yxdXdY . Let's denote P=\frac{dY}{dX}P=dXdY

Then the equation can be rewritten in form

X(y-x^2\frac{P}{y})=y\frac{x^2}{y^2}P^2X(y−x2yP)=yy2x2P2

By multiplying both sides with y, we assume

X(y^2-x^2P)=x^2P^2X(y2−x2P)=x2P2

Or

X(Y-XP)=XP^2X(Y−XP)=XP2

Therefore

Y=XP+P^2Y=XP+P2

which is now in Clairaut’s form

The solution got by just replacing P by constant c.

Hence

Y=cX+c^2Y=cX+c2

or

y^2=cx^2+c^2y2=cx2+c2

Explanation:

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