Differential Equations (VSAQs)
Maths-2B | 8. Differential Equations – VSAQs:
Welcome to VSAQs in Chapter 8: Differential Equations. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.
VSAQ-1 : Find the order and degree of the differential equation [d2 y/dx2 +(dy/dx)3]6/5 = 6y
Given D.E is $$(\frac{d^2y}{dx^2} + (\frac{dy}{dx})^3)^\frac{6}{5} = 6y$$
$$\Rightarrow \frac{d^2y}{dx^2} + (\frac{dy}{dx})^3 = (6y)^\frac{5}{6}$$
Here, the highest order derivative is $$\frac{d^2y}{dx^2}$$ and the exponent of $$\frac{d^2y}{dx^2}$$ is 1
VSAQ-2 : Find the order and degree of the D.E x1∕2 (d2y/dx2)1⁄3 + x dy/dx + y = 0
Given D.E is $$x^{\frac{1}{2}}\left(\frac{d^2y}{dx^2}\right)^{\frac{1}{3}} + x \frac{dy}{dx} + y = 0$$
$$\Rightarrow x^{\frac{1}{2}}\left(\frac{d^2y}{dx^2}\right)^{\frac{1}{3}} = -(x \frac{dy}{dx} + y)$$
Cubing on both sides, we get $$x^{\frac{3}{2}} \frac{d^2y}{dx^2} = -\left(x \frac{dy}{dx} + y\right)^3$$
Here, the highest order derivative is $$\frac{d^2y}{dx^2}$$
The exponent of $$\frac{d^2y}{dx^2}$$ is 1
VSAQ-3 : Find the order and degree of the differential equation d2 y/dx2 = [1 + (dy/dx)2]5∕3
Given D.E is $$\frac{d^2y}{dx^2} = [1 + (\frac{dy}{dx})^2]^\frac{5}{3}$$
Cubing both sides, we get $$\left(\frac{d^2y}{dx^2}\right)^3 = [1 + (\frac{dy}{dx})^2]^5$$
This form is free from fractional powers.
Here, the highest derivative in the equation after modification is $$\left(\frac{d^2y}{dx^2}\right)^3$$
For the given D.E, the order is 2 and the degree, which is the exponent of the highest derivative after the modification, is 3
VSAQ-4 : Form the differential equation corresponding to y = A cos 3x + B sin 3x,where A and B are parameters
Given $$y = A \cos 3x + B \sin 3x$$
$$\Rightarrow \frac{dy}{dx} = -3A(\sin 3x) + 3B(\cos 3x)$$
$$\Rightarrow \frac{d^2y}{dx^2} = -9A \cos 3x – 9B \sin 3x = -9(A \cos 3x + B \sin 3x) = -9y$$
$$\frac{d^2y}{dx^2} + 9y = 0$$
VSAQ-5 : Form the D.E corresponding to y = cx – 2c2 where c is a parameter
Given that $$y = cx – 2c^2$$
Differentiating w.r.to x we get $$y’ = c$$
From (1) $$y = y’x – 2y’^2$$
$$\Rightarrow 2y’^2 – xy’ + y = 0$$
$$\Rightarrow 2\left(\frac{dy}{dx}\right)^2 – x\left(\frac{dy}{dx}\right) + y = 0$$
VSAQ-6 : Find the order of the differential equation of the family of all circles with their centres at the origin
The equation of the family of circles with centers at the origin is $$x^2 + y^2 = r^2$$
Differentiating w.r.to x, we get $$2x + 2y \frac{dy}{dx} = 0$$
$$\Rightarrow x + y \frac{dy}{dx} = 0$$
VSAQ-7 : Find the D.E corresponding to y = acos (nx+b) where a,b are parameters
Given equation $$y = a\cos(nx + b)$$
On differentiating (1) w.r.t x we get $$y’ = -a\sin(nx + b)n$$
Differentiating (2) w.r.t x we get $$y” = -an^2\cos(nx + b)$$
$$\Rightarrow y” = -n^2a\cos(nx + b)$$
$$\Rightarrow y” + n^2y = 0$$
$$\Rightarrow y” + n^2y = 0$$
VSAQ-8 : Find the general solution of dy/dx = ex+y
Given D.E is $$\frac{dy}{dx} = e^{x+y} = e^x \cdot e^y$$
$$\Rightarrow \frac{dy}{e^y} = e^x dx$$
$$\Rightarrow e^{-y} dy = e^x dx$$
$$\Rightarrow e^x dx – e^{-y} dy = 0$$
Hence $$\int e^x dx – \int e^{-y} dy = c$$
$$\Rightarrow e^x + e^{-y} = c$$
VSAQ-9 : Solve dy/dx + 1 = ex+y
Given D.E is $$\frac{dy}{dx} + 1 = e^{x+y}$$
Put x + y = t
Then $$1 + \frac{dy}{dx} = \frac{dt}{dx}$$
From (1) $$\Rightarrow \frac{dt}{dx} = e^t$$
$$\Rightarrow \frac{dt}{e^t} = dx$$
$$\Rightarrow \int \frac{dt}{e^t} = \int dx$$
$$\Rightarrow -e^{-t} = x + c$$
$$\Rightarrow e^{-(x + y)} + x + c = 0$$
VSAQ-10 : Solve dy/dx = 1+y2/1+x2
Given D.E is $$\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}$$
$$\Rightarrow \frac{dy}{1 + y^2} = \frac{dx}{1 + x^2}$$
$$\Rightarrow \int \frac{dy}{1 + y^2} = \int \frac{dx}{1 + x^2}$$
$$\Rightarrow \tan^{-1}y = \tan^{-1}x + c$$
The solution is $$\tan^{-1}y = \tan^{-1}x + c$$
VSAQ-11 : Solve √1 – x2 dy + √1 – y2 dx = 0
Given D.E is $$\sqrt{1 – x^2} \, dy + \sqrt{1 – y^2} \, dx = 0$$
$$\Rightarrow \sqrt{1 – x^2} \, dy = -\sqrt{1 – y^2} \, dx$$
$$\Rightarrow \frac{dy}{\sqrt{1 – y^2}} = -\frac{dx}{\sqrt{1 – x^2}}$$
$$\Rightarrow \int \frac{dy}{\sqrt{1 – y^2}} = -\int \frac{dx}{\sqrt{1 – x^2}}$$
$$\Rightarrow \sin^{-1}y = -\sin^{-1}x + c$$
The solution is $$\sin^{-1}y + \sin^{-1}x = c$$
VSAQ-12 : Find the general solution of x + y dy/dx = 0
Given D.E is $$x + y \frac{dy}{dx} = 0$$
$$\Rightarrow y \frac{dy}{dx} = -x$$
$$\Rightarrow ydy = -xdx$$
$$\Rightarrow \int ydy = – \int xdx$$
$$\Rightarrow \frac{y^2}{2} = – \frac{x^2}{2} + c$$
The general solution is $$y^2 + x^2 = 2c$$
VSAQ-13 : Find the general solution of dy/dx = 2y/x
Given D.E is $$\frac{dy}{dx} = \frac{2y}{x}$$
$$\Rightarrow \frac{dy}{y} = 2\frac{dx}{x}$$
$$\Rightarrow \int \frac{dy}{y} = 2\int \frac{dx}{x}$$
$$\Rightarrow \log y = 2 \log x + \log c$$
$$\Rightarrow \log y = \log x^2 + \log c$$
$$\Rightarrow \log y = \log c x^2$$
$$\Rightarrow y = c x^2$$
The general solution is $$y = c x^2$$
VSAQ-14 : Find the I.F of the D.E (cosx) dy/dx + ysinx = tanx by transforming it into linear form
Given D.E is $$(\cos x) \frac{dy}{dx} + y \sin x = \tan x$$
$$\Rightarrow \frac{dy}{dx} + y\left(\frac{\sin x}{\cos x}\right) = \frac{\tan x}{\cos x}$$
$$\Rightarrow \frac{dy}{dx} + y \tan x = \tan x \sec x$$
This is in the form $$\frac{dy}{dx} + yP(x) = Q(x)$$
So it is a linear D.E in y.
Here $$P(x) = \tan x$$
I.F (Integrating Factor) = $$e^{\int P(x)dx} = e^{\int \tan x dx} = e^{\log \sec x} = \sec x$$
VSAQ-15 : Find the I.F of x dy/dx – y = 2x2 sec2 2x by transforming it into linear form
Given D.E is $$x \frac{dy}{dx} – y = 2x^2 \sec^2(2x)$$
$$\Rightarrow \frac{dy}{dx} – \frac{y}{x} = 2x \sec^2(2x)$$
This is in the form $$\frac{dy}{dx} + yP(x) = Q(x)$$
So, it is a linear D.E in y.
Here $$P(x) = -\frac{1}{x}$$
Integrating Factor (I.F) = $$e^{\int P(x)dx} = e^{\int -\frac{1}{x} dx} = e^{-\log x} = e^{\log \frac{1}{x}} = \frac{1}{x}$$