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

Solve the following differential equations:

Question 1.

Solution:

This is a Homogeneous differential equation

Seperating the variables

Question 2.

(x^{3} + y^{3}) dy – x^{2}y dx = 0

Solution:

(x^{3} + y^{3}) dy – x^{2}y dx = 0

(x^{3} + y^{3}) dy = x^{3}y dx

This is a Homogeneous differential equation.

Separating the variables

Question 3.

Solution:

This is a Homogeneous differential equation

Seperating the variables

Question 4.

2xydx + (x^{2} + 2y^{2}) dy = 0

Solution:

2xy dx + (x^{2} + 2y^{2}) dy = 0

(x^{2} + 2y^{2}) dy = – 2xy dx

This is a Homogeneous differential equation

Separating the variables

Question 5.

(y^{2} – 2xy)dx = (x^{2} – 2xy)dy

Solution:

(y^{2} – 2xy) dx = (x^{2} – 2xy) dy

This is a Homogeneous differential equation

Seperating the variables

Question 6.

Solution:

Separating the variables

Question 7.

Solution:

This is a Homogeneous differential equation

Separating the variables

Question 8.

(x^{2} + y^{2}) dy = xy dx. It is given that y(1) = 1 and y(x_{0}) = e. Find the value of x_{0}.

Solution:

This is a Homogeneous differential equation

Seperating the variables

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

Question 1.

Solve: (2\(\sqrt{x y}\))dy + y dx = 0

Solution:

Question 2.

Solve: (x^{3} + 3xy^{2})dx + (y^{3} + 3x^{2}y)dy = 0

Solution:

Integrating, we have

Question 3.

Solution:

Which is in terms of v alone.

⇒ the given problem comes under homogeneous type.

Question 4.

Solve: (x^{2} + y^{2})dy = xy dx

Solution:

Which is a function in v alone.

⇒ the given problem comes under homogeneous type.

Question 5.

Find the equation of the curve passing through (1, 0) and which has slope \(1+\frac{y}{x}\) at (x, y).

Solution:

Given the curve passes through (1, 0) ⇒ at x = 1, y = 0

0 = log 1 +c ⇒ c = 0