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Question 6. Let ∗ be the binary operation on N given by a ∗ b = L.C.M. of a and b. Find

(v) Which elements of N are invertible for the operation ∗?

test

Q 15)
$f(x)\left\{\begin{matrix} 2x & if &x<0 \\ 0& if &0\leq x\leq 1 \\ 4x&if & x>1 \end{matrix}\right.$

Given function is Given function is satisfies for the all real values of x case (i)  k < 0 Hence, function is continuous for all values of x < 0 case (ii)  x = 0 L.H.L at x= 0 R.H.L. at x = 0 L.H.L. = R.H.L. = f(0) Hence, function is continuous at x = 0 case (iii)  k > 0 Hence , function is continuous for all values of x > 0 case (iv) k < 1  Hence , function is...

Q18.    The general solution of the differential equation $e^x dy + (y e^x + 2x) dx = 0$ is

(A)    $xe^y + x^2 = C$

(B)    $xe^y + y^2 = C$

(C)    $ye^x + x^2 = C$

(D)    $ye^y + x^2 = C$

Given equation is we can rewrite it as It is  type of equation where  Now, Now, the general solution is Therefore, (C) is the correct answer

Q16.    The general solution of the differential equation $\frac{ydx - xdy}{y} = 0$is

(A)    $xy = C$

(B)    $x = Cy^2$

(C)    $y = Cx$

(D)    $y = Cx^2$

Given equation is we can rewrite it as  Integrate both the sides we will get Therefore, answer is (C)

Q17.    The general solution of a differential equation of the type $\frac{dx}{dy} + P_1 x = Q_1$ is

(A)    $ye^{\int P_1 dy} = \int \left(Q_1 e^{\int P_1 dy} \right )dy +C$

(B)    $ye^{\int P_1 dx} = \int \left(Q_1 e^{\int P_1 dx} \right )dx +C$

(C)    $xe^{\int P_1 dy} = \int \left(Q_1 e^{\int P_1 dy} \right )dy +C$

(D)    $xe^{\int P_1 dx} = \int \left(Q_1 e^{\int P_1 dx} \right )dx +C$

Given equation   is and we know that the general equation of such type of differential equation is Therefore, the correct answer is (C)

Q15.    The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20, 000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?

Let n be the population of the village at any time t. According to question, Now, at t=0, n = 20000 (Year 1999) Again, at t=5, n= 25000 (Year 2004) Using these values, at t =10 (Year 2009) Therefore, the population of the village in 2009 will be 31250.

Q14.    Find a particular solution of the differential equation $(x+1)\frac{dy}{dx} = 2e^{-y} -1$, given that $y = 0$ when $x = 0$

Given equation is we can rewrite it as Integrate both the sides Put   put   again Put this in our equation Now, by using boundary conditions we will find the value of C It is given that  y = 0 when x = 0 at   x = 0 Now, put the value of C Therefore, the particular solution is

Q13.    Find a particular solution of the differential equation $\frac{dy}{dx} + y \cot x = 4x \textup{cosec} x\ (x\neq 0)$, given that $y = 0 \ \textup{when}\ x = \frac{\pi}{2}$.

Given equation is This is    type where  and  Now,                       Now, the solution of given differential equation is given by relation Now, by using boundary conditions we will find the value of C It is given that  y = 0 when  at    Now, put the value of C Therefore, the particular solution is

Q12.    Solve the differential equation $\left[\frac{e^{-2\sqrt x}}{\sqrt x} - \frac{y}{\sqrt x} \right ]\frac{dx}{dy} = 1\; \ (x\neq 0)$.

Given, This is equation is in the form of    p =   and Q =  Now, I.F. =  We know that the solution of the given differential equation is:

Q11.     Find a particular solution of the differential equation $(x - y) (dx + dy) = dx - dy,$ , given that$y = -1$, when $x = 0$. (Hint: put $x - y = t$)

Given equation is Now, integrate both the sides Put Now, given equation become Now, integrate both the sides Put   again Now, by using boundary conditions we will find the value of C It is given that y = -1 when x = 0 Now, put the value of C Therefore,  the particular solution of the differential equation    is

Q10.    Solve the differential equation $ye^\frac{x}{y}dx = \left(e^\frac{x}{y} + y^2 \right )dy\ (y \neq 0)$

Given, Let  Differentiating it w.r.t. y, we get, Thus from these two equations,we get,

Q8.    Find the equation of the curve passing through the point $\left(0,\frac{\pi}{4} \right )$ whose differential equation is $\sin x \cos y dx + \cos x \sin y dy = 0.$

Given equation is we can rewrite it as Integrate both the sides Now by using boundary conditiond, we will find the value of C It is given that the curve passing through the point  So, Now, Therefore, the equation of the curve passing through the point  whose differential equation is  is

Q9.    Find the particular solution of the differential equation $(1 + e^ {2x} ) dy + (1 + y^2 ) e^x dx = 0$, given that $y = 1$ when $x = 0$.

Given equation is we can rewrite it as Now, integrate both the sides Put   Put  again Put this in our equation Now, by using boundary conditions we will find the value of C It is given that y = 1 when x = 0 Now, put the value of C Therefore,  the particular solution of the differential equation    is

Q7.    Show that the general solution of the differential equation $\frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0$ is given by $(x + y + 1) = A (1 - x - y - 2xy)$, where A is parameter.

Given, Integrating both sides, Let   Let  A = ,  Hence proved.

Q6.    Find the general solution of the differential equation $\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1-x^2}} = 0$

Given equation is we can rewrite it as Now, integrate on both the sides Therefore, the general solution of the differential equation    is

Q5.    Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

Now, equation of the circle with center at (x,y) and radius r is Since, it  touch the coordinate axes in first quadrant  Therefore, x = y = r                      -(i) Differentiate it w.r.t x we will get           -(ii) Put value from equation (ii) in equation (i) Therefore,  the differential equation of the family of circles in the first quadrant which touches the coordinate axes is

Q4.    Prove that $x^2 - y^2 = c (x^2 + y^2 )^2$ is the general solution of differential equation$(x^3 - 3x y^2 ) dx = (y^3 - 3x^2 y) dy$, where c is a parameter.

Given, Now, let y = vx Substituting the values of y and y' in the equation,  Integrating both sides we get,   Now,   Let        Now,  Let v2 = p Now, substituting the values of I1 and I2 in the above equation, we get, Thus,

Q3.    Form the differential equation representing the family of curves given by $(x-a)^2 + 2y^2 = a^2$ , where a is an arbitrary constant.

Given equation is                     we can rewrite it as                    -(i) Differentiate both the sides w.r.t x                                             -(ii) Put value from equation (ii)  in (i)   Therefore, the required differential equation is

Q2.    Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(iv)    $x^2 = 2y^2\log y\qquad : \ (x^2 + y^2)\frac{dy}{dx} - xy = 0$

Given, Now, differentiating both sides w.r.t. x, Putting  values in LHS Therefore, the given function is the solution of the corresponding differential equation.

Q2.    Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(iii)    $y= x\sin 3x \qquad : \ \frac{d^2y}{dx^2} + 9y - 6\cos 3x = 0$

Given, Now, differentiating both sides w.r.t. x, Again, differentiating both sides w.r.t. x, Therefore, the given function is the solution of the corresponding differential equation.
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