OP Malhotra Class 12 Maths Solutions Chapter 17 Differential Equations Ex 17(e)

Parents can use ISC Class 12 OP Malhotra Solutions Chapter 17 Differential Equations Ex 17(e) to provide additional support to their children.

S Chand Class 12 ICSE Maths Solutions Chapter 17 Differential Equations Ex 17(e)

Solve the following differential equations:

Question 1.
2xy \(\frac { dy }{ dx }\) = x² + y²
Solution:
Given 2xy\(\frac { dy }{ dx }\) = x² + y²

Question 2.
x²\(\frac { dy }{ dx }\) = 2xy + y²
Solution:

Question 3.
x²\(\frac { dy }{ dx }\) = x² + 5xy + 4y²
Solution:

Question 4.
x²\(\frac { dy }{ dx }\) = y(x + y)
Solution:

Question 5.
y² + x²\(\frac { dy }{ dx }\) = xy\(\frac { dy }{ dx }\)
Solution:

Question 6.
x²\(\frac { dy }{ dx }\) = (x² – 2y² + xy)
Solution:

Question 7.
x²y dx = (x³ + y³)dy
Solution:

Question 8.
\(\frac { dy }{ dx }\) = \(\frac { y }{ x }\) + tan\(\frac { y }{ x }\)
Solution:
Given \(\frac{d y}{d x}=\frac{y}{x}+\tan \frac{y}{x}=\phi\left(\frac{y}{x}\right)\) … (1)
which is homogeneous differential equation.
put y = vx ⇒ \(\frac { dy }{ dx }\) = v + x\(\frac { dv }{ dx }\)
∴ from eqn. (1); we have
v + x \(\frac { dv }{ dx }\) = v + tan v ⇒ x\(\frac { dv }{ dx }\) = tan v
⇒ \(\frac { dv }{ tan v }\) = \(\frac { dx }{ x }\) ; on integratmg ; we have
∫ cot v dv = ∫\(\frac { dx }{ x }\) + logc
⇒ log | sin v | = log | x | + log c
⇒ sin v = cx ⇒ sin (\(\frac { y }{ x }\)) = cx
be the required solution.

Question 9.
\(\left[x \sqrt{x^2+y^2}-y^2\right]\)dx + xy dy = 0
Solution:
Given differential equation can be written as :

Question 10.
(x² – 3xy²)dx = (y³ – 3x²y)dy
Solution:
Given diff. eqn. can be written as \(\frac{d y}{d x}=\frac{x^3-3 x y^2}{y^3-3 x^2 y}\) … (1)
which is homogeneous diff. eqn.

Question 11.
\(\frac{d y}{d x}=\frac{y}{x}+\sin \frac{y}{x}\)
Solution:
Given \(\frac{d y}{d x}=\frac{y}{x}+\sin \frac{y}{x}\) … (1)
which is homogeneous diff. eqn.
put y = vx ⇒ \(\frac { dy }{ dx }\) = v + x\(\frac { dv }{ dx }\)
∴ from (1); we have

which is the required solution.

Question 12.
x(\(\frac { dy }{ dx }\)) = y(log y – log x + 1)
Solution:
Given diff. eqn. can be written as ;

Question 13.
x² dy + y (x + y) dx = 0, given that y = 1 when x = 1.
Solution:
Given diff. eqn. can be written as ;
x²dy + y(x + y)dx = 0

be the required solution.

Question 14.
\(\frac{d y}{d x}=\frac{x y}{x^2+y^2}\) given that y = 1 when x = 0.
Solution:
Given diff. eqn. be,
\(\frac{d y}{d x}=\frac{x y}{x^2+y^2}\) … (1)
which is homogeneous diff. eqn.
put y = vx ⇒ \(\frac { dy }{ dx }\) = v + x\(\frac { dv }{ dx }\)
∴ from (1) ; we have

given that y = 1, when x = 0
∴ from (2) ; we have
0 + log 1 = c ⇒ x = 0
Thus eqn. (2) becomes ;
– \(\frac{x^2}{2 y^2}\) ⇒ log y = \(\frac{x^2}{2 y^2}\)
⇒ y = \(e^{\frac{x^2}{2 y^2}}\)
be the required particular solution.

Question 15.
Find the particular solution of the differential equation 2ye^x/ydx + (y – 2xe^x/y)dy = 0 given that x = 0, when y = 1.
Solution:
Given diff. eqn. can be written as,

given that x = 0 when y = 1
∴ from (2); we have 2 = c
∴ from (2); we have 2e^x/y = – log | y| + 2
⇒ e^x/y = – \(\frac { 1 }{ 2 }\) log|y| + 1
which is the required solution.