What is the degree of the differential equation representing the family of curves \(y^{2}=\sqrt{c}\,x\), where \(c\) is a positive parameter?
1
Writing \(k=\sqrt{c}\), the curve is \(y^{2}=kx\); differentiating gives \(2y y'=k\). Substituting back, \(y^{2}=2xy y'\), i.e. \(y=2x y'\), which is a first-degree equation in \(y'\), so the degree is \(1\).
Consider the following in respect of the differential equation:
\(\frac{{{d^2}y}}{{d{x^2}}} + 2{\left( {\frac{{dy}}{{dx}}} \right)^2} + 9y = x\)
1. The degree of the differential equation is 1.
2. The order of the differential equation is 2.
Which of the above statements is/are correct?
The degree of the differential equation \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} - {\rm{x}} = {\left( {{\rm{y}} - {\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^{ - 4}}\) is
What is the degree of the differential equation? \(1+\left(\frac{dy}{dx}\right)^2 =\left(\frac{d^2y}{dx^2}\right)^{\frac{4}{3}}?\)
What is the order of the differential equation of all ellipses whose axes are along the coordinate axes?
What is the degree of the differential equation of all circles touching both the coordinate axes in the first quadrant?
What is the degree of the differential equation \(\frac{{{d}^{3}}y}{d{{x}^{3}}}+{{\left( \frac{dy}{dx} \right)}^{2}}-{{x}^{2}}\left( \frac{{{d}^{4}}y}{d{{x}^{4}}} \right)=0?\)
The differential equation of the family of curves y = p cos (ax) + q sin (ax), where p, q are arbitrary constants, is
The order and degree of the differential equation y 2= 4a (x – a), where ‘a’ is an arbitrary constant, are respectively
Consider the following statements :
1. The degree of the differential equation \(\frac{\text{dy}}{\text{dx}} + \cos \left(\frac{\text{dy}}{\text{dx}}\right)\) = 0 is 1.
2. The order of the differential equation \(\left(\frac{\text{d}^2\text{y}}{\text{dx}^2}\right)^3 + \cos \left(\frac{\text{dy}}{\text{dx}}\right)\) = 0 is 2.
Which of the statements given above is/are correct?
What are the order and degree, respectively, of the differential equation \({\left( {\frac{{{{\rm{d}}^3}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^3}}}} \right)^2} = {{\rm{y}}^4} + {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^5}?\)
Consider the following in respect of the differential equation:
\(\frac{{{d^2}y}}{{d{x^2}}} + 2{\left( {\frac{{dy}}{{dx}}} \right)^2} + 9y = x\)
1. The degree of the differential equation is 1.
2. The order of the differential equation is 2.
Which of the above statements is/are correct?
The partial differential equation \(\frac{{\partial u}}{{\partial t}} + u\frac{{\partial u}}{{\partial x}} = \frac{{{\partial ^2}u}}{{\partial {x^2}}}\) is a
The degree of the differential equation \({\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^3} + {\left( {\frac{{dy}}{{dx}}} \right)^2} + \sin x\left( {\frac{{dy}}{{dx}}} \right) + y = 0\) is:
In the following partial differential equation, θ is a function of t and z, and D and K are functions of θ
\(D\left( \theta \right)\frac{{{\delta ^2}\theta }}{{\delta {z^2}}} + \frac{{\delta K\left( \theta \right)}}{{\delta z}} - \frac{{\delta \theta }}{{\delta t}} = 0\)
The above equation isThe solution of the equation \({\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{y}} = 0{\rm{}}\) passing through the point (1,1) is