The solution lof d y d x − x t a n ( y − x ) = 1 Options: A . &nbsp c e e x 2 2 = s i n ( y − x ) B . &nbsp c e e x 2 2 = s i n ( y + x ) C . &nbsp c e e x 2 2 = s i n ( y − x ) 2 D . &nbsp c e e x 2 2 = c o s ( y − x )

Answer: Option A : A Put y-x = z. Then d y d x − 1 = d z d x ⇒ d y d x = 1 + d z d x Given d y d x − x t a n ( y − x ) = 1 ⇒ 1 + d z d x − x t a n z = 1 ⇒ d z d x = x t a n z ⇒ 1 t a n z d z = x d x ⇒ ∫ c o t z d z = ∫ x d x ⇒ l o g | s i n z | − l o g c = x 2 2 ⇒ l o g | s i n z c | = x 2 2 = s i n z c = e x 2 2 ⇒ s i n ( y − x ) = c e x 2 2 ∴ The solution is s i n ( y − x ) = c e x 2 2 , where c is arbitrary constant.

The solution lof dydx−x tan(y−x)=1

Discussion Forum : Differential Equations

Question -

The solution lof $\frac{d y}{d x} - x t a n (y - x) = 1$

Options:

A .

c e \frac{e^{x^{2}}}{2} = s i n (y - x)

B .

c e \frac{e^{x^{2}}}{2} = s i n (y + x)

C .

c e \frac{e^{x^{2}}}{2} = s i n (y - x)^{2}

D .

c e \frac{e^{x^{2}}}{2} = c o s (y - x)

Answer: Option A
:
A
Put y-x = z. Then

\frac{d y}{d x} - 1 = \frac{d z}{d x} \Rightarrow \frac{d y}{d x} = 1 + \frac{d z}{d x}

Given

\frac{d y}{d x} - x t a n (y - x) = 1 \Rightarrow 1 + \frac{d z}{d x} - x t a n z = 1 \Rightarrow \frac{d z}{d x} = x t a n z \Rightarrow \frac{1}{t a n z} d z = x d x \Rightarrow \int c o t z d z = \int x d x

\Rightarrow l o g | s i n z | - l o g c = \frac{x^{2}}{2} \Rightarrow l o g | \frac{s i n z}{c} | = \frac{x^{2}}{2} = \frac{s i n z}{c} = \frac{e^{x^{2}}}{2} \Rightarrow s i n (y - x) = c \frac{e^{x^{2}}}{2}

∴

The solution is

s i n (y - x) = c \frac{e^{x^{2}}}{2}

, where c is arbitrary constant.

Was this answer helpful ?

Next Question

Discussion Forum : Differential Equations

Submit Your Solution hear: