Test: First Order Differential Equations (non-linear) - Computer Science Engineering (CSE) MCQ

Dividing both sides by z (logz)²

Integrating Factor

The given differential equation is

xD² – 2y D + 4x = 0

The discriminant of this equation is

(-2y)² – 4x.4x = 0 ⇒ y² – 4x² = 0

Differentiating with respect to x,

⇒ yD = 4x ⇒ D = 4x/y

Substituting D = 4x/y

⇒ y² – 4x² = 0

Hence y² = 4x² satisfies the given differential equation. The singular solution is

y² – 4x² = 0 ⇒ y = ±2x

dy/dx = p
p² = tan⁡(y − xp)
tan⁻¹p² = y − xp
y = xp + tan⁻¹p²

y = px + f(p)

Which is clairaut’s equation.

So equation is:
y = cx + tan⁻¹c²

Integrating factor, IF = e^∫dx = e^x

y(IF) = ∫(IF)e^xdx + C

At x = 0, y = 1
∴ C = 1/2

Equation of circle having centre at (r, 0) and radius r is written as:

(x - r)² + (y - 0)² = r²

x² + r² - 2xr + y² = r²

x² - 2xr + y² = 0

x² + y² = 2xr        ___(1)

Differentiating on both sides:

x + yy’ = r

Put value of r in equation (1):

Thus, x² + y² = 2 (x + yy’) x

2xyy’ + x² - y² = 0

Note:

A differential equation is said to be linear if the dependent variable and its derivative appears in first degree.

Note:

 is the first order linear differential equation.

The two criterions for linearity of an equation are:

• The dependent variable y and its derivatives are of first degree.
• Each coefficient depends only on the independent variable

Error Function is given by, 
Some of its properties are:
erf (0) = 0
erf (∞) = 1
erf (-x) = -erf(x)