Formation of Differential Equations

1 Mark Questions

Write the degree of the differential equation

<math display="block">\left(\frac{dy}{dx}\right)^4 + 3y\frac{d^2y}{dx^2} = 0.</math>

Delhi 2013C

The degree of the differential equation is the degree of the highest order derivative, when differential coefficients are made free from radicals and fractions sign.

Given differential equation is

<math display="block">\left(\frac{dy}{dy}\right)^4 + 3y\frac{d^2y}{dy^2} = 0</math>

Here, highest order derivative is <math>d^2y/dx^2</math>, whose degree is one. So, degree of differential equation is 1. (1)

1. Write the degree of the differential equation

<math display="block">x^3 \left(\frac{d^2y}{dv^2}\right)^2 + x \left(\frac{dy}{dv}\right)^4 = 0.</math>

Delhi 2013

Given differential equation is

<math display="block">x^{3} \left( \frac{d^{2}y}{dy^{2}} \right)^{2} + x \left( \frac{dy}{dy} \right)^{4} = 0</math>

Here, all differential coefficients are free from radical sign.

<math>\therefore</math> Degree = 2 (1) Write the degree of the differential equation

<math>\left(\frac{dy}{dy}\right)^4 + 3x\frac{d^2y}{dx^2} = 0.</math>

Delhi 2013

Given differential equation is

<math>\left(\frac{dy}{dx}\right)^{2} + 3x\left(\frac{d^{2}y}{dx^{2}}\right) = 0</math> (ux

Here, all differential coefficients are free from radical sign. (1) ∴ Degree = 1

1. Write the differential equation representing the family of curves <math>y = mx</math>, where m is an

arbitrary constant. All India 2013

Given, family of curves is <math>y = mx</math> ...(i)

where, m is an arbitrary constant.

Now, differentiating Eq. (i) w.r.t. x, we get

<math>\frac{dy}{dx} = m</math>

On putting <math>m = \frac{dy}{dx}</math> in Eq. (i), we get

<math>y = x \frac{dy}{dx}</math>

which is the required differential equation. (1)

4 Marks Questions

1. Find the differential equation of the family of circles in the first quadrant which touch the

coordinate axes. All India 2010C