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Question: Write the differential equation representing the family of curves \(y=mx\), where m is an arbitrary ...

Write the differential equation representing the family of curves y=mxy=mx, where m is an arbitrary constant

Explanation

Solution

As we are given an equation of a curve y=mxy=mx in the question, where m is an arbitrary constant. We need to find the differential equation represented by the equation of the given curve y=mxy=mx. So, firstly we need to find dydx\dfrac{dy}{dx} by differentiating the equation with respect to x and thereby find the value of m. Later, find the relation between x, y, and dydx\dfrac{dy}{dx} to get the differential equation.

Complete step-by-step solution:
We have been given an equation of a curve as y=mx......(1)y=mx......(1)
We have to find the value of the term “m” in the above equation first. We know that “m” indicates the slope of the curve. We also know that the derivative of y with respect to x is another way of representing the slope of a curve.
So, differentiate the equation (1) with respect to x, then we have:
dydx=m......(2)\dfrac{dy}{dx}=m......(2)
Hence, we get the value of m in equation (2). Now, substitute the value of m in equation (1), we get:
y=dydxx......(3)y=\dfrac{dy}{dx}x......(3)
Now, solve the equation (3) to get the final differential equation as:
xdydxy=0x\dfrac{dy}{dx}-y=0
Hence, xdydxy=0x\dfrac{dy}{dx}-y=0 is the required differential equation.

Note: There is another method to solve the given equation of curve y=mxy=mx to find the differential equation. We can get the value of m directly from the equation of curve instead of differentiating it first as:
y=mx m=yx......(1) \begin{aligned} & y=mx \\\ & \Rightarrow m=\dfrac{y}{x}......(1) \\\ \end{aligned}
Now, differentiate the equation (1) with respect to x on both sides, by applying the difference rule of differentiation, i.e. ddx(f(x)g(x))=(f(x).g(x)f(x).g(x)(g(x))2)\dfrac{d}{dx}\left( \dfrac{f(x)}{g(x)} \right)=\left( \dfrac{f'(x).g(x)-f(x).g'(x)}{{{\left( g(x) \right)}^{2}}} \right)
So, we get:
0=x.dydxy.1x2 0=xdydxy......(2) \begin{aligned} & \Rightarrow 0=\dfrac{x.\dfrac{dy}{dx}-y.1}{{{x}^{2}}} \\\ & \Rightarrow 0=x\dfrac{dy}{dx}-y......(2) \\\ \end{aligned}
Hence, equation (2) is the required differential equation of given curve y=mxy=mx.
Be careful while applying the difference rule of differentiation. Do not differentiate the numerator separately and denominator separately. That is wrong method. Always, apply the difference rule formula correctly, i.e. ddx(f(x)g(x))=(f(x).g(x)f(x).g(x)(g(x))2)\dfrac{d}{dx}\left( \dfrac{f(x)}{g(x)} \right)=\left( \dfrac{f'(x).g(x)-f(x).g'(x)}{{{\left( g(x) \right)}^{2}}} \right).