93. Find the equation of the curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangents to the curve at that point by 5.

4 Views|Posted 8 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Vishal Baghel | Contributor-Level 10

8 months ago

We know that slope of tangent to the curve is $\frac{d y}{d x}$

$∴ x + y = \frac{d y}{d x} + 5$

$\frac{d y}{d x} - y = x - 5$ Which has form $\frac{d y}{d x} + P x = Q$

$w h e r e, P = 1 & Q = x - 5$

$∴ I . F = e^{\int P d x} = e^{\int - 1 d x} = e^{- x}$

Thus the solution has the form

$\begin{array}{l} y e^{- x} = \int (x - 5) e^{- x} d x + c \\ = \int x e^{- x} d x - \int 5 e^{- x} d x + c \\ = y e^{- x} = I + 5 e^{- x} + c \\ w h e r e, I = \int x e^{- x} d x \\ = x \int e^{- x} d x - \int \frac{d}{d x} x \int e^{- x} d x d x \\ = - x e^{- x} + \int e^{- x} d x \\ = - x e^{- x} - e^{- x} \end{array}$

$\begin{array}{l} ∴ y e^{- x} = - x e^{- x} - e^{- x} + 5 e^{- x} + c \\ = y e^{- x} = - x e^{- x} + 4 e^{- x} + c \\ = y = - x + 4 + \frac{c}{e^{- x}} \\ = y + x = 4 + c e^{x} \end{array}$

Given, the curve passes through (0,2) so y=2 when x=0

$\begin{array}{l} 2 + 0 = 4 c e^{0} \\ 2 - 4 = c \\ c = - 2 \end{array}$

$∴$ The particular solution is

$y + x = 4 - 2 e^{x}$

Upvote

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 = $\frac{1}{2}$

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

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 $x, y = - \frac{1}{8}$

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

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.