Solve x^2 y d x- (x^3+y^3 ) d y=0 - Vidyadip

Question

Solve $x^2 y d x-\left(x^3+y^3\right) d y=0$

✓

Answer

$
\begin{aligned}
& x^2 y d x-\left(x^3+y^3\right) d y=0 \\
& \therefore x^2 y d x-\left(x^3+y^3\right)=d y \\
& \therefore \frac{d y}{d x}=\frac{x^2 y}{x^3+y^3} \ldots \text { (i) }
\end{aligned}
$
Put $y=t x$
Differentiating w.r.t. $x$, we get
$
\frac{d y}{d x}=t+x \frac{d t}{d x}
$
Substituting (ii) and (iii) in (i), we get
$
\begin{aligned}
& t+x \frac{d t}{d x}=\frac{x^2 \cdot t x}{x^3+t^3 x^3} \\
& \therefore t+x \frac{d t}{d x}=\frac{x^3 \cdot t}{x^3\left(1+t^3\right)} \\
& \therefore x \frac{d t}{d x}=\frac{t}{1+t^3}-t \\
& \therefore x \frac{d t}{d x}=\frac{t-t-t^4}{1+t^3} \\
& \therefore x \frac{d t}{d x}=\frac{-t^4}{1+t^3} \\
& \therefore \frac{1+t^3}{t^4} d t=-\frac{d x}{x}
\end{aligned}
$
Integrating on both sides, we get
$
\begin{aligned}
& \int \frac{1+t^3}{t^4} d t=-\int \frac{1}{x} d x \\
& \therefore \int\left(\frac{1}{t^4}+\frac{1}{t}\right) d t=-\int \frac{1}{x} d x \\
& \therefore \int t^{-4} d t+\int \frac{1}{t} d t=-\int \frac{1}{x} d x \\
& \therefore \frac{t^3}{-3}+\log |t|=-\log |x|+c \\
& \therefore-\frac{1}{3 t^3}+\log |t|=-\log |x|+c \\
& \therefore-\frac{1}{3} \cdot \frac{1}{\left(\frac{y}{x}\right)^3}+\log \left|\frac{y}{x}\right|=-\log |x|+c \\
& \therefore \frac{x^3}{3 y^3}+\log |y|-\log |x|=-\log |x|+c \\
& \therefore \log |y|-\frac{x^3}{3 y^3}=c
\end{aligned}
$

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