Advertisements
Advertisements
प्रश्न
Form the differential equation of the family of curves represented by y2 = (x − c)3.
Advertisements
उत्तर
The equation of the family of curves is \[y^2 = \left( x - c \right)^3\] ...(1)
where \[c \in R\] is a parameter.
This equation contains only one parameter, so we shall obtain a differential equation of first order.
Differentiating equation (1) with respect to \[x\] , we get
\[2y\frac{dy}{dx} = 3 \left( x - c \right)^2\] ...(2)
\[\frac{y^2}{2y\frac{dy}{dx}} = \frac{\left( x - c \right)^3}{3 \left( x - c \right)^2}\]
\[ \Rightarrow \frac{y}{2\frac{dy}{dx}} = \frac{\left( x - c \right)}{3}\]
\[ \Rightarrow \frac{3y}{2\frac{dy}{dx}} = x - c\]
\[ \Rightarrow c = x - \frac{3y}{2\frac{dy}{dx}}\]
Substituting the value of \[c\] in equation (1), we get
\[y^2 = \left( x - x + \frac{3y}{2\frac{dy}{dx}} \right)^3 \]
\[ \Rightarrow y^2 = \frac{27 y^3}{8 \left( \frac{dy}{dx} \right)^3}\]
\[ \Rightarrow 8 y^2 \left( \frac{dy}{dx} \right)^3 = 27 y^3 \]
\[ \Rightarrow 8 \left( \frac{dy}{dx} \right)^3 - 27y = 0\]
\[\frac{y^2}{2y\frac{dy}{dx}} = \frac{\left( x - c \right)^3}{3 \left( x - c \right)^2}\]
\[ \Rightarrow \frac{y}{2\frac{dy}{dx}} = \frac{\left( x - c \right)}{3}\]
\[ \Rightarrow \frac{3y}{2\frac{dy}{dx}} = x - c\]
\[ \Rightarrow c = x - \frac{3y}{2\frac{dy}{dx}}\]
Substituting the value of \[c\] in equation (1), we get
\[y^2 = \left( x - x + \frac{3y}{2\frac{dy}{dx}} \right)^3 \]
\[ \Rightarrow y^2 = \frac{27 y^3}{8 \left( \frac{dy}{dx} \right)^3}\]
\[ \Rightarrow 8 y^2 \left( \frac{dy}{dx} \right)^3 = 27 y^3 \]
\[ \Rightarrow 8 \left( \frac{dy}{dx} \right)^3 - 27y = 0\]
It is the required differential equation.
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
