Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(b)

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Differential Equations Solutions Exercise 8(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Differential Equations Solutions Exercise 8(b)

I.

Question 1.
Find the general solution of \(\sqrt{1-x^2}\) dy + \(\sqrt{1-y^2}\) dx = 0.
Solution:
Given differential equation is
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(b) 1
sin-1 y = – sin-1 x + c
Solution is sin-1 x + sin-1 y = c, where c is a constant.

Question 2.
Find the general solution of \(\frac{dy}{dx}=\frac{2y}{x}\).
Solution:
\(\frac{dy}{dx}=\frac{2y}{x}\)
∫\(\frac{dy}{dx}\) = 2∫\(\frac{2y}{x}\)
log c + log y = 2 log x
log cy = log x²
Solution is cy = x², where c. is a constant.

II. Solve the following differential equations.

Question 1.
\(\frac{dy}{dx}=\frac{1+y^2}{1+x^2}\)
Solution:
\(\frac{dy}{dx}=\frac{1+y^2}{1+x^2}\)
∫\(\frac{dy}{1+y^2}\) = ∫\(\frac{dx}{1+x^2}\)
tan-1 y = tan-1 x + tan-1c where c is a constant.

Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(b)

Question 2.
\(\frac{dy}{dx}\) = ey-k
Solution:
\(\frac{dy}{dx}=\frac{e^y}{e^x}\)
\(\frac{dy}{e^y}=\frac{dx}{e^x}\)
∫e-xdx = ∫e-ydy
-e-x = -e-y + C
e-y = e-x + c where c is a constant.

Question 3.
(ex + 1) y dy + (y + 1) dx = 0
Solution:
(ex + 1 )y. dy = – (y + 1) dx
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(b) 2
y – log (y + 1) = log (e-x + 1) + log c
⇒ y – log (y + 1) = log c (e-x + 1)
⇒ y = log (y + 1) + log c (e-x + 1)
y = log c (y + 1) (e-x +1)
Solution is
ey = c(y + 1) (e-x +1)

Question 4.
\(\frac{dy}{dx}\) = ex-y + x²e-y
Solution:
\(\frac{dy}{dx}\) = ex-y + x² . e-y
= \(\frac{e^x}{e^y}=\frac{x^2}{e^y}\)
∫ey . dy = ∫(ex + x²) dx
Solution is
ey = ex + \(\frac{x^3}{3}\) + c

Question 5.
tan y dx + tan x dy = 0
Solution:
tan y dx = – tan x dy
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(b) 3
log sin x = – log sin y + log c
log sin x + log sin y = log c
log (sin x . sin y) = log c
⇒ sin x . sin y = c is the solution

Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(b)

Question 6.
\(\sqrt{1+x^2}\)dx + \(\sqrt{1+y^2}\)dy = 0
Solution:
\(\sqrt{1+x^2}\)dx = –\(\sqrt{1+y^2}\)dy
Integrating both sides we get
∫\(\sqrt{1+x^2}\)dx = -∫\(\sqrt{1+y^2}\)dy
Integrating both sides we get
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(b) 4

Question 7.
y – x\(\frac{dy}{dx}\) = 5(y² + \(\frac{dy}{dx}\))
Solution:
y – 5y² = (x + 5)\(\frac{dy}{dx}\)
\(\frac{dx}{x+5}=\frac{dy}{y(1-5y}\)
Integrating both sides
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(b) 5

Question 8.
\(\frac{dy}{dx}=\frac{xy+y}{xy+x}\)
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(b) 6

III. Solve the following differential equations.

Question 1.
\(\frac{dy}{dx}=\frac{1+y^2}{(1+x^2)xy}\)
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(b) 7
log (1 + y²) = log x² – log (1 + x²) + log c
log (1 + x²) + log (1 + y²) = log x² + log c
Solution is (1 + x²) (1 + y²) = cx²

Question 2.
\(\frac{dy}{dx}\) + x² = x² e3y
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(b) 8
log(1 – e-3y) = x³ + c'(c’ = 3c)
Solution is
1 – e-3y = e . k(k = ec’)

Question 3.
(xy² + x)dx+(yx²+y)dy = 0.
Solution:
(xy² + x) dx + (yx² + y) dy = 0
x(y² + 1) dx + y (x² + 1) dy = 0
Dividing with (1 + x²) (1 + y²)
\(\frac{x dx}{1+x^2}+\frac{y dy}{1+y^2}\) = 0
Integrating
∫\(\frac{x dx}{1+x^2}\) + ∫\(\frac{y dy}{1+y^2}\) = 0
\(\frac{1}{2}\) [(log (1 + x²) + log (1 + y²)] = log c
log (1 + x²) (1 + y²) = 2 log c = log c²
Solution is (1 + x²) (1 + y²) = k when k = c².

Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(b)

Question 4.
\(\frac{dy}{dx}\) = 2y tanh x
Solution:
\(\frac{dy}{dx}\) = 2y tanh x
\(\frac{dy}{y}\) = 2 tanh x dx
Integrating both sides we get
∫\(\frac{dy}{y}\) = 2 ∫ tanh x dx
log y = 2 log |cosh x| + log c
lny = 2ln cosh x + In c
y = c cos²h x

Question 5.
sin-1 \(\frac{dy}{dx}\) = x + y
Solution:
\(\frac{dy}{dx}\) = sin(x + y)
x + y = t
1 + \(\frac{dy}{dx}=\frac{dt}{dx}\)
\(\frac{dt}{dx}\) – 1 = sin t
\(\frac{dt}{dx}\) = 1 + sin t
Integrating both sides we get
∫\(\frac{dt}{1+\sin t}\) = ∫dx
∫\(\frac{1-\sin t}{\cos^2 t}\) dt = x + c
∫sec² t dt – ∫tan t . sec t dt = x + c
tan t – sec t = x + c
⇒ tan (x + y) – sec (x + y) = x + c

Question 6.
\(\frac{dy}{dx}+\frac{y^2+y+1}{x^2+x+1}\) = 0
Solution:
\(\frac{-dy}{y^2+y+1}=\frac{dx}{x^2+x+1}\)
Integrating both sides dy
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(b) 9

Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(b)

Question 7.
\(\frac{dy}{dx}\) = tan² (x + y)
Solution:
\(\frac{dy}{dx}\) = tan² (x + y)
put v = x + y
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(b) 10
2v + sin 2v = 4x + c’
2(x + y) + sin 2(x + y) = 4x + c’
x – y – \(\frac{1}{2}\)sin [2(x + y)] = c

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