Differential Equations 2 Question 1
1. The general solution of the differential equation $\left(y^{2}-x^{3}\right) d x-x y d y=0(x \neq 0)$ is (where, $C$ is a constant of integration)
(2019 Main, 12 April II)
(a) $y^{2}-2 x^{2}+C x^{3}=0$
(b) $y^{2}+2 x^{3}+C x^{2}=0$
(c) $y^{2}+2 x^{2}+C x^{3}=0$
(d) $y^{2}-2 x^{3}+C x^{2}=0$
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Answer:
Correct Answer: 1. (b)
Solution:
- Given differential equation is
$$ \begin{array}{cc} & \left(y^{2}-x^{3}\right) d x-x y d y=0,(x \neq 0) \\ \Rightarrow \quad & x y \frac{d y}{d x}-y^{2}=-x^{3} \end{array} $$
Now, put $y^{2}=t \Rightarrow 2 y \frac{d y}{d x}=\frac{d t}{d x} \Rightarrow y \frac{d y}{d x}=\frac{1}{2} \frac{d t}{d x}$
$\therefore \quad \frac{x}{2} \frac{d t}{d x}-t=-x^{3}$
$\Rightarrow \quad \frac{d t}{d x}-\frac{2}{x} t=-2 x^{2}$
which is the linear differential equation of the form
$$ \frac{d t}{d x}+P t=Q $$
Here, $\quad P=-\frac{2}{x}$ and $Q=-2 x^{2}$.
Now, $\quad$ IF $=e^{-\int \frac{2}{x} d x}=\frac{1}{x^{2}}$
$\because$ Solution of the linear differential equation is (IF) $t=\int Q$ (IF) $d x+\lambda \quad$ [where $\lambda$ is integrating constant]
$$ \begin{aligned} & \therefore \quad t \frac{1}{x^{2}}=-2 \int x^{2} \times \frac{1}{x^{2}} d x+\lambda \\ & \Rightarrow \quad \frac{t}{x^{2}}=-2 x+\lambda \\ & \Rightarrow \quad \frac{y^{2}}{x^{2}}+2 x=\lambda \quad\left[\because t=y^{2}\right] \\ & \Rightarrow \quad y^{2}+2 x^{3}-\lambda x^{2}=0 \\ & \text { or } \quad y^{2}+2 x^{3}+C x^{2}=0 \quad[\operatorname{let} C=-\lambda] \end{aligned} $$
