Ask & Answer Question

Differential Equations Ncert Solutions Maths class 12th Maths Class 12th Differential Equations

108. Find the particular solution of the differential equation $\frac{d y}{d x} + y c o t x = 4 x c o s e c x (x \neq 0)$

given that y = 0 when x = π/2

2 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

The given differential equation is:

$\frac{d y}{d x} + y c o t x = 4 x c o s e c x$

This equation is a linear equation of the form

$\begin{array}{l} \frac{d y}{d x} + P y = Q, w h e r e, p = c o t x & Q = 4 x c o s e c x \\ N o w, I . F = e^{\int P d x} = e^{\int c o t x d x} = e^{l o g | s i n x |} = s i n x \end{array}$

The general solution of the given differential equation is given by,

$y (I . F) = \int (Q \times I . F .) d x + C$

$\begin{array}{l} \Rightarrow y s i n x = \int (4 x c o s e c x . s i n x) d x + C \\ \Rightarrow y s i n x = 4 \int x d x + C \\ \Rightarrow y s i n x = 4 . \frac{x^{2}}{2} + C \\ \Rightarrow y s i n x = 2 x^{2} + C . . . . . . . . . . (1) \\ N o w, y = 0 a t, x = \frac{π}{2} \end{array}$

Therefore, equation (1) becomes:

$\begin{array}{l} 0 = 2 \times \frac{π}{2} + C \\ \Rightarrow C = - \frac{π^{2}}{2} \end{array}$

Substituting $C = - \frac{π^{2}}{2}$ in equation (1), we get:

$y s i n x = 2 x^{2} - \frac{π^{2}}{2}$

This is the required particular solution of the given differential equation.

0 Upvotes 0 Downvotes

Similar Questions for you

Let α be the angle between the lines whose direction cosines satisfy the equations I + m – n = 0 and I² + m² – n² = 0. Then the value of sin⁴α + cos⁴α is:

Vishal Baghel

l + m – n = 0

l + m = n . (i)

l² + m² = n²

Now from (i)

l² + m² = (l + m)²

=> 2lm = 0

=>lm = 0

l = 0 or m = 0

=> m = n Þ l = n

if we take direction consine of line

$0, \frac{1}{\sqrt[]{2}}, \frac{1}{\sqrt[]{2}} a n d \frac{1}{\sqrt[]{2}}, 0, \frac{1}{\sqrt[]{2}}$

cos a = $\frac{1}{2}$

$s i n^{4} α + c o s^{4} α = {(\frac{\sqrt[]{3}}{2})}^{4} = {(\frac{1}{2})}^{4} = \frac{9}{16} + \frac{1}{16} = \frac{10}{16} = \frac{5}{8}$

If y = y(x) is the solution of the differential equation

$(1 + e^{2 x}) \frac{d y}{d x} + 2 (1 + y^{2}) e^{x} = 0 a n d y (0) = 0,$ then $6 (y' (0) + {(y (l o g_{e} \sqrt[]{3}))}^{2})$ is equal to

Vishal Baghel

$\Rightarrow \int_{}^{} \frac{d y}{1 + y^{2}} = - 2 \int_{}^{} \frac{e^{x} d x}{1 + {(e^{x})}^{2}} + C$

$\Rightarrow t a n^{- 1} y = - 2 \cdot t a n^{- 1} e^{x} + C$

x = 0, y = 0

$\Rightarrow 0 = 2 t a n^{- 1} + C$

$C = + \frac{π}{2}$

now at x = $l n \sqrt[]{3}$

$t a n^{- 1} y = - 2 t a n^{- 1} (e^{l n \sqrt[]{3}}) + \frac{π}{2}$

$6 (y' (0) + {(y (l n \sqrt[]{3}))}^{2}) = 6 (- 1 + \frac{1}{3}) = - 4$

If x₁ $\underset{x \to \infty}{l i m} \frac{c o t^{- 1} (\sqrt[]{x + 1} - \sqrt[]{x})}{s e c^{- 1} ({(\frac{2 x + 1}{x - 1})}^{x})}$ then $\frac{(3 e^{x_{1}} + x_{2})}{4}$ equals

Vishal Baghel

$x_{1} = \underset{x \to \infty}{l i m} \frac{2^{\frac{x^{n}}{e^{x}}} - 3^{\frac{x^{n}}{e^{x}}}}{\frac{x^{n}}{e^{x}}}, p u t \frac{x^{n}}{e^{x}} = t$

$x_{2} = \underset{x \to \infty}{l i m} \frac{c o t^{- 1} (\sqrt[]{x + 1} - \sqrt[]{x})}{s e c^{- 1} ({(\frac{2 x + 1}{x - 1})}^{x})}$

$x_{2} = \frac{2}{π} \underset{x \to \infty}{l i m} t a n^{- 1} (\sqrt[]{x + 1} + \sqrt[]{x})$

$x_{2} = \frac{2}{π} . \frac{π}{2} = 1$

Let $y = y (x)$ be a function of $x$ satisfying $y \sqrt[]{1 - x^{2}} = k - x \sqrt[]{1 - y^{2}}$ where $k$ is a constant and $y (\frac{1}{2}) = - \frac{1}{4}$ . Then $\frac{d y}{d x}$ at $x = \frac{1}{2}$ , is equal to:

alok kumar singh

Differentiating

y. $\frac{- 2 x}{2 \sqrt[]{1 - x^{2}}} + \sqrt[]{1 - x^{2}} \cdot y^{'} = \frac{x \cdot 2 y y^{'}}{2 \sqrt[]{1 - y^{2}}} - \sqrt[]{1 - y^{2}}$

Put $x = \frac{1}{2}, y = - \frac{1}{4}$ and $x, y = - \frac{1}{8}$

$y^{'} = \frac{- \sqrt[]{5}}{2}$

Let a curve y = f(x) pass through the point (2, (logₑ2)²) and have slope 2y/(xlogₑx) for all positive real value of x. Then the value of f(e) is equal to……….

alok kumar singh

dy/dx = 2y/ (xlnx).
dy/y = 2dx/ (xlnx).
ln|y| = 2ln|lnx| + C.
ln|y| = ln (lnx)²) + C.
y = A (lnx)².
(ln2)² = A (ln2)². ⇒ A=1.
y = f (x) = (lnx)².
f (e) = (lne)² = 1² = 1.

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

108. Find the particular solution of the differential equation $\frac{d y}{d x} + y c o t x = 4 x c o s e c x (x \neq 0)$

given that y = 0 when x = π/2

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

108. Find the particular solution of the differential equation dydx+ycotx=4xcosecx(x≠0) given that y = 0 when x = π/2

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

108. Find the particular solution of the differential equation $\frac{d y}{d x} + y c o t x = 4 x c o s e c x (x \neq 0)$

given that y = 0 when x = π/2