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Let a solution y = y(x) of the differential equation
STATEMENT1 :
STATEMENT2 :
The given differential equation is
⇒ Statement 1 is true.
The order and degree of the differential equation
The solution of the equation
The degree and order of the differential equation of the family of all parabolas whose axis is x  axis, are respectively.
y^{2} = 4a (x  h), 2 yy_{1} = 4a ⇒ yy_{1} = 2a Differentiating, ⇒ y_{1}^{2} + yy_{2} = 0
Degree = 1, order = 2.
The solution of the differential equation
The differential equation for the family of circle x^{2} + y^{2}  2ay = 0, where a is anarbitrary constant is
Solution of the differential equation ydx + (x + x^{2} y)dy = 0 is
ydx + (x + x^{2}y)dy = 0
It is Bernoullis form. Divide by x^{2}
It is linear in t.
The differential equation representing the family of curves is a parameter, is of order and degree as follows :
[On putting value of c from (ii) in (i)]
On simplifying, we get (y  2 xy') 2 = 4 yy'^{3} ........ (iii)
Hence equation (iii) is of order 1 and degree 3.
then the solution of theequation is
The differential equation whose solution is Ax^{2} + By^{2} = 1 where A and B are arbitrary constants is of
.........(1)
..........(2)
........(3)
From (2) and (3)
Dividing both sides by –B, we get
Which is a DE of order 2 and degree 1.
The differential equation of all circles passing through the origin and having their centres on the xaxis is
General equation of circles passing through origin and having their centres on the xaxis is
x^{2} + y^{2} + 2gx = 0...(i)
On differentiating w.r.t x, we get
∴ equation (i) be
The soluton of the differential equation satisfying the condition y(1) =1 is
The differential equation which represents the family of curves are arbitrary constants, is
Solution of the differential equation
cos x dy = y(sin x y) dx
From equation (i)
then y (ln 2) is equal to :
Let I be the purchase value of an equipment and V (t) be the value after it has been used for t years. The value V(t) depreciates at a rate given by differential equation where k > 0 is a constant and T is thetotal life in years of the equipment. Then the scrap value V(T) of the equipment is
The population p (t) at time t of a certain mouse species satisfies the differential equation If p (0) = 850, then the time at which the population becomes zero is :
Given differential equation is
Integrate both sides, we get
∴ t_{1} = 2ln 18
At present, a firm is manufacturing 2000 items. It is estimated that the rate of change of production P w.r.t. additional number of workers x is given by If thefirm employs 25 more workers, then the new level of production of items is
By integrating
Given, when x = 0 then P = 2000
⇒ C = 2000
Now when x = 25 then
P = 100 × 25 – 8 × (25)3/2 + 2000
= 4500 – 1000
⇒ P = 3500
Let the population of rabbits surviving at time t be governed by the differential equation If p(0) = 100, then p(t) equals:
Given differential equation is
By separating the variable, we get
Integrating on both the sides,
Using given condition p(t) = 400 – 300 e^{t/2}
Let y(x) be the solution of the differential equation Then y (e) is equal to:
y logx = 2[x log x – x] + c
Put x = 1, y.0 = –2 + c ⇒ c = 2
Put x = e
y loge = 2e(log e – 1) + c ⇒ y(e) = c = 2
If a curve y = f(x) passes through the point (1, –1) and satisfies the differential equation, y(1 + xy) dx = x dy, then
y (1 + xy)dx = xdy
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