find the equation of the curve which passes through the origin and has the slope x + 3y - 1 at any point (x, y) on it. - methods of solving first order, first degree differential equations - linear differential equations
find the equation of the curve which passes through the origin and has the slope x + 3y - 1 at any point (x, y) on it.
let a (x, y) be the point on the curve y = f(x). then slope of the tangent to the curve at point a is `"dy"/"dx"`. according to the given condition, `"dy"/"dx" = "x" + 3"y" - 1` ∴ `"dy"/"dx" - 3"y" = "x - 1"` ....(1) this is the linear differential equation of the form `"dy"/"dx" + "py" = "q"`, where p = - 3 and q = x - 1 ∴ i.f. = `"e"^(int "p dx") = "e"^(int - 3"dx") = "e"^(- 3"x")` ∴ the solution of (1) is given by `"y" * ("i.f.") = int "q" * ("i.f.") "dx" + "c"` ∴ `"y" * "e"^(- 3"x") = int ("x - 1") * "e"^(-3"x") "dx" + "c"` ∴ `"e"^(- 3"x") * "y" = ("x - 1") int "e"^(- 3"x") - int ["d"/"dx" ("x - 1") * int "e"^(- 3"x") "dx"] "dx" + "c"\_1` ∴ `"e"^(- 3"x") * "y" = ("x - 1") * "e"^(- 3"x")/-3 - int 1 * "e"^(- 3"x") * "y"/-3 "dx" + "c"\_1` ∴ `"e"^(- 3"x") * "y" = - 1/3 ("x - 1") * "e"^(- 3"x") + 1/3 int "e"^(- 3"x") "dx" + "c"\_1` ∴ `"e"^(- 3"x") * "y" = - 1/3 ("x - 1") "e"^(- 3"x") + 1/3 * "e"^(- 3"x")/-3 + "c"\_1` ∴ `"e"^(- 3"x") * "y" = - 1/3 ("x - 1")"e"^(- 3"x") - 1/9 "e"^(- 3"x") + "c"\_1` ∴ 9y = - 3(x - 1) - 1 + 9`"c"\_1 * e^("3x")` ∴ 9y + 3(x - 1) + 1 = `9"c"\_1 * e^("3x")` ∴ 9y + 3x - 3 + 1 = `9"c"\_1 * e^("3x")` ∴ 3(x + 3y) = 2 + `9"c"\_1 * e^("3x")` ∴ 3(x + 3y) = 2 + `"c" * e^("3x")` where c = 9c1 ....(2) this is the general equation of the curve. but the required curve is passing through the origin (0, 0). ∴ by putting x = 0 and y= 0 in (2), we get 0 = 2 + c ∴ c = - 2 ∴ from (2), the equation of the required curve is 3(x + 3y) = 2 - 2`e^("3x")` i.e. 3(x + 3y) = 2(1 - `e^("3x")`).
slope of the line ⇒ slope of the lineequation of line :⇒ 3(y - 5) = 2(x - 2)⇒ 2x - 3y + 11 = 0 …equation (i)putting x = x + 3 and y = y + 2 in the above equation we get2(x + 3) - 3(y + 2) + 11 = 0⇒ 2x - 3y + 11 = 0since it again gives the same equation as (i) so it is a point on this line.hence proved y = 3x

It can be observed that x = −1, y = −3 and x = 1, y = 3 are solutions of the above equation. Therefore, the solution table is as follows.



| x | -1 | 1 |
| y | -3 | 3 |

The graph of the above equation is constructed as follows.

![](https://www.sarthaks.com/?qa=blob&qa_blobid=15681428413536225923) switch x and y, then solve again for y:

x = 3y+12
x-12 = 3y
(x-12)/3 = y

or,

y = x/3 - 4 hint: assume the co-ordinates of $r$ as $(x,y)$ and as it is given that $p,q,r$ are collinear points. apply distance condition of collinear points which is if $p,q,r$ are three collinear points then they must satisfy $pq+qr=pr$ and get one equation. apply distance formula which states that if $a({{x}\_{1}},{{y}\_{1}})$ and $b({{x}\_{2}},{{y}\_{2}})$ are two points then the distance $ab=\sqrt{{{({{x}\_{2}}-{{x}\_{1}})}^{2}}+{{({{y}\_{2}}-{{y}\_{1}})}^{2}}}$ and get the second equation by using $pr=10$ , solve these two equations and get the co-ordinates of $r$. complete step-by-step solution:it is given in the question that $p, q, r$ are collinear points.collinear means lying on the same line which means that $p, q, r$ lies on the same straight line.let us suppose that the co-ordinates of $r$ is $(x,y)$. now, we have $p(7,7)$, $q(3,4)$, $r(x,y)$ and also $pr=10$.we have to use distance formula in order to find $pq,qr,pr$. so, according to distance formula if $a({{x}\_{1}},{{y}\_{1}})$ and $b({{x}\_{2}},{{y}\_{2}})$ then $ab=\sqrt{{{({{x}\_{2}}-{{x}\_{1}})}^{2}}+{{({{y}\_{2}}-{{y}\_{1}})}^{2}}}$.now, $p,q,r$ are three collinear points then they must satisfy,$\therefore pq+qr=pr$using distance formula, we get$\rightarrow \sqrt{{{(7-3)}^{2}}+{{(7-4)}^{2}}}+\sqrt{{{(3-x)}^{2}}+{{(4-y)}^{2}}}=10$$\rightarrow \sqrt{16+9}+\sqrt{{{(3-x)}^{2}}+{{(4-y)}^{2}}}=10$$\rightarrow 5+\sqrt{{{(3-x)}^{2}}+{{(4-y)}^{2}}}=10$\[\rightarrow \sqrt{{{(3-x)}^{2}}+{{(4-y)}^{2}}}=5\]squaring both sides, we get$\rightarrow {{(3-x)}^{2}}+{{(4-y)}^{2}}=25$$\rightarrow {{x}^{2}}+{{y}^{2}}-6x-8y+25=25$cancelling 25 from both sides, we get$\rightarrow {{x}^{2}}+{{y}^{2}}=6x+8y...................(1)$also, it is given that $pr=10$, using this we get$\therefore pr=10$\[\rightarrow \sqrt{{{(7-x)}^{2}}+{{(7-y)}^{2}}}=10\]squaring both sides, we get\[\rightarrow {{(7-x)}^{2}}+{{(7-y)}^{2}}=100\]$\rightarrow {{x}^{2}}+{{y}^{2}}-14x-14y+98=100$putting the value of ${{x}^{2}}+{{y}^{2}}$ from equation (1), we get$\rightarrow 6x+8y-14x-14y=100-98$$\rightarrow -8x-6y=2$now, expressing $y$ in terms of $x$, we get$\rightarrow y=\dfrac{-(2+8x)}{6}=\dfrac{-(1+4x)}{3}..................(2)$putting the value of $y$ from equation (2) in equation (1), we get$\rightarrow {{x}^{2}}+{{\left( \dfrac{-(1+4x)}{3} \right)}^{2}}=6x+\dfrac{-(1+4x)}{3}\times 8$$\rightarrow \dfrac{9{{x}^{2}}+1+8x+16{{x}^{2}}}{9}=\dfrac{18x-8-32x}{3}$$\rightarrow 25{{x}^{2}}+8x+1=3(-14x-8)$$\rightarrow 25{{x}^{2}}+8x+1+42x+24=0$$\rightarrow 25{{x}^{2}}+50x+25=0$$\rightarrow 25({{x}^{2}}+2x+1)=0$$\rightarrow {{(x+1)}^{2}}=0$$\rightarrow x+1=0\rightarrow x=-1$now, putting the value of $x$ in equation (2), we get$\rightarrow y=\dfrac{-(1+4\times -1)}{3}$$\rightarrow y=\dfrac{3}{3}=1$so, $x=-1$ and $y=1$.hence the co-ordinates of $r$ is $(-1,1)$. note: this question is simple and there are many other alternative ways to solve it but the tricky part is to choose which method to solve the question. there are three different conditions of collinearity as there is slope condition, distance condition, and area condition. as in the other two conditions distances are not involved, only the distance condition involves the use of distances. in the question, distance is mentioned so we choose that method to solve it. hence, students should carefully read the question and choose which method is suitable to solve the question according to the values given. the graph of this equation is a line.  we know that because each term is either a constant (1.5) or the product of a constant and the first power of a variable (d, 0.8t).
since we have two variables (d and t), the line sits in a two-dimensional plane, with a vertical axis and a horizontal axis.  d stands alone in the equation, so its value depends on the value oft.  therefore, the vertical--or dependent--axis will represent values of d, and the horizontal--or independent--axis will represent values of t.
at this point, then, we can set up a coordinate plane with a vertical axis labeled d and a horizontal axis labeled t.  all of the points on our plane have a coordinate (t, d).
to graph a line, we need two points.  since the question asks for the axial intercepts anyway, we can use those.
we'll start with the d-intercept, where the line crosses the d-axis.  all of the points on the d-axis have a t-coordinate of zero, so to find where the line crosses the axis, we substitute zero for t in our equation and solve.
d = 1.5 + 0.8 × t
d = 1.5 + 0.8 × (0)
d = 1.5 + 0
d = 1.5
so our d-intercept has the coordinates (0, 1.5), and we have our first point and our first intercept.
similarly, to find the t-intercept, we note that all of the points on the t-axis have a d-coordinate of zero, so we set d equal to zero and solve.
d = 1.5 + 0.8 × t
0 = 1.5 + 0.8 × t
0 - 1.5 = 1.5 - 1.5 + 0.8 × t
-1.5 = 0.8 × t
-1.5 ÷ 0.8 = (0.8 ÷ 0.8) × t
-1.875 = 1 × t
-1.875 = t
and the d-intercept is (-1.875, 0).
now we have both axial intercepts, so that part of the question is complete.  we can plot those two points and draw the line that passing through them (extending on, at either side, to infinity).
all that remains is to find the turning points.  a turning point, also called a local maximum or local minimum, is where the value of a function goes from decreasing to increasing or from increasing to decreasing.  (the graph shown at http://en.wikipedia.org/wiki/file:parabola\_with\_focus\_and\_directrix.svg has a minimum at the origin, for example.)
a line has a constant slope.  to say that differently, the angle between any two points on a line is the same as the angle between any two other points on the line.  if a line is increasing in one location (like our line), it is increasing everywhere, and it is increasing everywhere at the same rate.  therefore, a line has no turning points. ?solution 
step 1:
we have to draw the graph of form r to r. 
step 2:
the graph of f is the set of all points (x,f(x)) with x in the domain of f.
we take x in [-4,4]
and we will find the separate values of y for separate intervals.
 
 
 

 
......(1)
 

 

 

using x values along x-axis and along y-axis we get the graph of given function form r to r.
step 3:
the graph is as follow given, x3 – 1, x – 1 as pair of polynomialsneed to find the quotient and remainder⇒ to find the quotient and remainder the given equation can be written as p(x) = (x – a)q(x) + b⇒ since, the polynomial is of third degree we can write the q(x) as x2 + ax + b∴ p(x) = (x – a)( x2 + ax + b) + c⇒ x3 + 1 = (x + 1)(x2 + ax + b) + c⇒ x3 + 1 = (x3 + x2 + ax2 + ax + bx + b) + c⇒ x3 + 1 = x3 + (a + 1)x2 + (b + a)x + (c + b)∴ a + 1 = 0, b + a = 0, c + b = 1⇒ a = – 1, b = 1⇒ c + b = 1⇒ c = 0quotient = x2 + ax + b = x2 – x + 1remainder = 0 (a) to my eye, they never have a common speed. that would require the two curves to have the same slope at the same time, which doesn't appear to me to be true.

(b) b is ahead of a but is slowing down and being passed by a, which has been moving at constant speed.

(c) b has an acceleration with greater magnitude. (in fact, the curve for a doesn't show any acceleration at all.)

(d) a has a constant slope, and therefore a constant velocity (and therefore displays no acceleration at all). b starts ahead of a initially with a velocity greater, but then decelerates as it nears t = 4 s. at around t = 3.7 s, b's velocity (slope) has nearly dropped to zero, and it gets passed by a.

if you find this helpful, please select favorite answer! general systems thursday, september 29, 2016 3:50 pm a fruit store sell two types of fruit, apples and oranges. apples are $2 and oranges are $3.today 50 fruit was sold and the store made $110. 110 = 2▯ + 3▯ 50 = ▯ + ▯ ▯ = 40 ▯ = 10 general ▯▯ ▯+ ▯ ▯▯ ▯⋯+ ▯ ▯ = ▯ ▯ a linear equation is an equation in the form of: • x's are variables and a's are constants. • solution set for a linear equation is the set of all solutions. • set of linear equations where a commonsolution is desired is called a system of linear equations. graphs • 1 variables is a point • 2 variables is a line • 3 variables is a plane • 4+ variables is a hyperplane examples of systems 2 equations, 2 unknowns • no solutions • one given : let a , b and c be three students whose chances of solving a problem is given i.e , p(a) = , p(b) = and p(c) = . p() = , p() = and p()= to find: the probability that the problem is solved .here , p(the problem is solved) = 1 - p(the problem is not solved)= 1- p( )= 1 – [p( ) p( ) p( )]= 1- []= 1 - = therefore , the probability that the problem is solved is .