Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 10 Ordinary Differential Equations Ex 10.6 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.6

Question 1.

Solve the following differential equations.

[x + y cos(\(\frac { y }{ x }\))] dx = x cos (\(\frac { y }{ x }\)) dy

Solution:

The given equation can be written as

On integration we obtain

which gives the required solution.

Question 2.

Solve (x³ + y³)dy – x² ydx = 0

Solution:

The given equation can be written as

Question 3.

Solve ye^{\(\frac { x }{ y }\)} dx = (x^{\(\frac { x }{ y }\)} + y)dy

Solution:

The given equation can be written as

Question 4.

Solve 2xy dx + (x² + 2y²)dy = 0

Solve ye^{\(\frac { x }{ y }\)} dx = (x^{\(\frac { x }{ y }\)} + y)dy

Solution:

The given differential equation can be written as

\(\frac { 1 }{ 3 }\) log (3v + 2v³) + log x = log |C_{1}|

log (3v + 2v³) + 3log (x) = 3 log (C_{1})

log (3v + 2v³) + log (x)³ = log (C_{1})³

log (3v + 2v³)x³ = log C_{1}³

(3v + 2v³)x³ = C_{1}³

3x²y + 2y³ = C_{1}³

3x²y + 2y³ = C is a required solution.

Question 5.

(y² – 2xy) dx = (x² – 2xy) dy

Solution:

Given equation is (y² – 2xy) dx = (x² – 2xy) dy

y² – 2xy = (x² – 2xy) \(\frac { dy }{ dx }\)

∴ The equation can written as

log (3v² – 3v) = – 3 log x + log C

log (3v² – 3v) = – log x³ + log C

= log c – log x³

Question 6.

x \(\frac { dy }{ dx }\) = y – x cos²(\(\frac { y }{ x }\))

Solution:

Given x \(\frac { dy }{ dx }\) = y – x cos² \(\frac { y }{ x }\)

The equation can be written as

\(\frac { dy }{ dx }\) = \(\frac { y-cos^2 \frac { y }{ x } }{ x }\) …….. (1)

This is a homogeneous differential equation.

y = vx

\(\frac { dy }{ dx }\) = v (1) + x \(\frac { dv }{ dx }\)

Substituting \(\frac { dy }{ dx }\) value in equation (1), we get

Integrating on both sides, we get

∫ sec² v dx = -∫ \(\frac { dx }{ x }\)

tan v = – log x + log C

tan v = log C – log x

tan v = log(\(\frac { C }{ x }\))

e^{tan v} = \(\frac { C }{ x }\)

C = x e^{tan v}

C = x e^{tan \(\frac { y }{ x }\)}

Is a required equation.

Question 7.

Solve (1 + 3e^{\(\frac { y }{ x }\)}) dy + 3e^{tan \(\frac { y }{ x }\)} (1 – \(\frac { y }{ x }\)) dx = 0, given that y = 0 when x = 1.

Solution:

The given differential equation may be

Given that y = 0 when x = 1

0 + 3(1) e° = c

3 = c

∴ y + 3xe^{y/x} = 3 is a required solution.

Question 8.

(x² + y²) dy = xy dx. It is given that y (1) = y(x_{0}) = e. Find the value of x_{0}.

Solution:

The given differential equation is of the form