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Differential Equation and Applications (p-1) — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Differential Equation and Applications (p-1)2 Marks
Question
Find the differential equation whose general solution is $x^3+y^3=$ 35ax.
✓

Answer

$
x^3+y^3=35 a x \ldots \text { (i) }
$
Differentiating w.r.t. $x$, we get
$
3 x^2+3 y^2 \frac{d y}{d x}=35 a
$
Substituting (ii) in (i), we get
$
x^3+y^3=\left(3 x^2+3 y^2 \frac{d y}{d x}\right) x
$
$\therefore x^3+y^3=3 x^3+3 x \cdot y^2 \frac{d y}{d x}$
$\therefore 2 x^3-y^3+3 x y^2 \frac{d y}{d x}=0$, which is the required differential equation.

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