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)5 Marks
Question
Solve $y^2 d x+\left(x y+x^2\right) d y=0$
✓
Answer
$ \begin{aligned} & y 2 d x+\left(x y+x^2\right) d y=0 \\ & \therefore\left(x y+x^2\right) d y=-y^2 d x \\ & \therefore \frac{d y}{d x}=-\frac{y^2}{x y+x^2} \ldots \end{aligned} $ Put $y=t x$...(ii) 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} & \therefore t+x \frac{d t}{d x}=\frac{-t^2 x^2}{x \cdot t x+x^2} \\ & \therefore t+x \frac{d t}{d x}=\frac{-t^2 x^2}{x^2(t+1)} \\ & \therefore x \frac{d t}{d x}=\frac{-t^2}{t+1}-t \\ & \therefore x \frac{d t}{d x}=\frac{-t^2-t^2-t}{t+1} \\ & \therefore x \frac{d t}{d x}=\frac{-\left(2 t^2+t\right)}{t+1} \\ & \therefore \frac{t+1}{2 t^2+t} d t=-\frac{1}{x} d x \end{aligned} $ Integrating on both sides, we get $ \begin{aligned} & \int \frac{t+1}{2 t^2+t} d t=-\int \frac{1}{x} d x \\ & \therefore \int \frac{2 t+1-t}{t(2 t+1)} d t=-\int \frac{1}{x} d x \\ & \therefore \int \frac{1}{t} d t-\int \frac{1}{2 t+1} d t=-\int \frac{1}{x} d x \\ & \therefore \log |t|-\frac{1}{2} \log |2 t+1|=-\log |x|+\log |c| \\ & \therefore 2 \log | t |-\log |2 t +1|=-2 \log | x |+2 \log | c | \\ & \therefore 2 \log \left|\frac{y}{x}\right|-\log \left|\frac{2 y}{x}+1\right|=-2 \log |x|+2 \log |c| \\ & \therefore 2 \log | y |-2 \log | x |-\log |2 y + x |+\log | x |=-2 \log | x |+2 \log | c | \\ & \therefore \log \left| y ^2\right|+\log | x |=\log \left| c ^2\right|+\log |2 y + x | \\ & \therefore \log \left| y ^2 x \right|=\log \left| c ^2( x +2 y )\right| \\ & \therefore xy ^2= c ^2( x +2 y ) \end{aligned} $
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