Question:
If \(l\) and \(m\) are order and degree of a differential equation of all the straight lines at constant distance of \(P\) units from the origin, then \(lm^2+l^2m=\)
If \(l\) and \(m\) are order and degree of a differential equation of all the straight lines at constant distance of \(P\) units from the origin, then \(lm^2+l^2m=\)
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For a family of straight lines at a fixed distance from the origin, use the distance formula \(\dfrac{|c|}{\sqrt{1+m^2}}\), then replace \(m\) by \(\dfrac{dy}{dx}\) and \(c\) by \(y-x\dfrac{dy}{dx}\).
Updated On: Jun 22, 2026
- \(2\)
- \(6\)
- \(12\)
- \(30\)
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The Correct Option is B
Solution and Explanation
Step 1: Write the general equation of a straight line.
Let the equation of a straight line be \[ y=mx+c \] where \(m\) is the slope and \(c\) is the intercept.
The distance of this line from the origin is \[ \frac{|c|}{\sqrt{1+m^2}} \] According to the question, this distance is \(P\). Hence, \[ \frac{|c|}{\sqrt{1+m^2}}=P \] Squaring both sides, \[ c^2=P^2(1+m^2) \]
Step 2: Convert into differential equation form.
From \[ y=mx+c, \] we have \[ m=\frac{dy}{dx} \] and \[ c=y-mx \] Therefore, \[ c=y-x\frac{dy}{dx} \] Substituting these values in \[ c^2=P^2(1+m^2), \] we get \[ \left(y-x\frac{dy}{dx}\right)^2 = P^2\left(1+\left(\frac{dy}{dx}\right)^2\right) \]
Step 3: Identify order and degree.
The highest order derivative present is \[ \frac{dy}{dx} \] So, the order is \[ l=1 \] The differential equation is polynomial in \(\frac{dy}{dx}\), and the highest power of \(\frac{dy}{dx}\) is \(2\).
So, the degree is \[ m=2 \]
Step 4: Calculate \(lm^2+l^2m\).
\[ lm^2+l^2m = 1(2)^2+(1)^2(2) \] \[ =4+2 \] \[ =6 \]
Step 5: Final conclusion.
Hence, \[ \boxed{6} \]
Let the equation of a straight line be \[ y=mx+c \] where \(m\) is the slope and \(c\) is the intercept.
The distance of this line from the origin is \[ \frac{|c|}{\sqrt{1+m^2}} \] According to the question, this distance is \(P\). Hence, \[ \frac{|c|}{\sqrt{1+m^2}}=P \] Squaring both sides, \[ c^2=P^2(1+m^2) \]
Step 2: Convert into differential equation form.
From \[ y=mx+c, \] we have \[ m=\frac{dy}{dx} \] and \[ c=y-mx \] Therefore, \[ c=y-x\frac{dy}{dx} \] Substituting these values in \[ c^2=P^2(1+m^2), \] we get \[ \left(y-x\frac{dy}{dx}\right)^2 = P^2\left(1+\left(\frac{dy}{dx}\right)^2\right) \]
Step 3: Identify order and degree.
The highest order derivative present is \[ \frac{dy}{dx} \] So, the order is \[ l=1 \] The differential equation is polynomial in \(\frac{dy}{dx}\), and the highest power of \(\frac{dy}{dx}\) is \(2\).
So, the degree is \[ m=2 \]
Step 4: Calculate \(lm^2+l^2m\).
\[ lm^2+l^2m = 1(2)^2+(1)^2(2) \] \[ =4+2 \] \[ =6 \]
Step 5: Final conclusion.
Hence, \[ \boxed{6} \]
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