Formation of a Differential Equation MCQ Quiz in मल्याळम - Objective Question with Answer for Formation of a Differential Equation - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Calculation:

Given, the circles pass through the origin.

They have their centres at (0, a) The circles have radius a.

∴ The equation of the family of circles in given by x² + (y - a)² = a² 

⇒ x2 + y² + a2 - 2ya = a2 

⇒ x2 + y² = 2ay ⋯ (i)

Differentiating wrt x, we get:

2x + 2y\(\frac{dy}{dx}\) = 2a\(\frac{dy}{dx}\)

⇒ a = \(\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}}\) ⋯ (ii)

Using (ii) in (i), we get:

x2 + y2 = 2y\(\left(\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}}\right)\) 

⇒ x2\(\frac{dy}{dx}\) + y2\(\frac{dy}{dx}\) = 2xy + 2y2\(\frac{dy}{dx}\)

⇒ (x2 - y2)​\(\frac{dy}{dx}\) - 2xy = 0

∴ The differential equation of all circles which pass through the origin and whose centres lie on y-axis is (x2 - y2)​\(\frac{dy}{dx}\) - 2xy = 0.

The correct answer is Option 1.

Calculation:

Given: y² – 2ay + x² = a²      ----(1)

Differentiating both sides w.r.t x

\( \Rightarrow \frac{{2{\rm{ydy}}}}{{{\rm{dx}}}} - 2{\rm{a}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + 2{\rm{x}} = 0\)

⇒ 2py – 2ap + 2x = 0

⇒ py + x = ap

\(\Rightarrow \frac{{{\rm{py}}\; + \;{\rm{x}}}}{{\rm{p}}} = {\rm{a}}\)

Put \({\rm{a}} = \frac{{{\rm{py}}\; + \;{\rm{x}}}}{{\rm{p}}}\) in equation (1)

\({{\rm{y}}^2} - 2{\rm{y}}\left( {\frac{{{\rm{py}} + {\rm{x}}}}{{\rm{p}}}} \right) + {{\rm{x}}^2} = {\left( {\frac{{{\rm{py}} + {\rm{x}}}}{{\rm{p}}}} \right)^2}\)

⇒ p²y² – 2py(py + x) + p²x² = (p²y² + x² + 2pxy)

⇒ – 2p²y² – 2pxy + p²x² – x² – 2pxy

⇒ (x² – 2y²)p² – 4pxy – x² = 0

Concept:

To form the differential equation of the given equation

• Differentiate the equation, the number of times as many as the constants are there.
• Find out the constants in terms of the variables.
• Substitute the variables in the original equation.

Calculation:

Given equation is \(\rm y^2+(x-b)^2 = c\)

There are two constants b and c so differentiate two times

Differentiating w.r.t x 

\(\rm 2y{dy\over dx}+2(x-b) = 0\)

\(\rm y{dy\over dx}=b-x\)

Differentiating again w.r.t x

\(\rm \left({dy\over dx}\right)^2+ y{d^2y\over dx^2}=-1\)

\(\rm y{d^2y\over dx^2}+\left({dy\over dx}\right)^2 +1 = 0\)

Concept:

To make a normal equation into the rational and integral form of derivative just eliminate the constants term from the equation by differentiating.

1. \(\rm \frac{de^x}{dx} = e^x\)

2. \(\rm \frac{de^{-x}}{dx} = -e^{-x}\)

Calculation:

Given that,

⇒ y = mex - ne-x       …. (1)

So, in this equation there is two variable m and n, so we have to remove them

Differentiate both sides with respect to x

⇒ \(\rm \frac {dy}{dx}\) = me^x + ne^-x

Again differentiate with respect to x

⇒ \(\rm \frac{d^2y}{dx^2}\) = mex - ne-x

So from equation 1;

⇒ \(\rm \frac{d^2y}{dx^2}\) = y

⇒ \(\rm \frac{d^2y}{dx^2}\) - y = 0

Given:

x = a(cos θ + θ sin θ)

y = a(sin θ - θ cos θ)

Solution:

\(\frac{dx}{dθ}\) = a(-sinθ + sinθ + θ cosθ)

= aθcosθ - (i)

\(\frac{dy}{dθ}\) = a(cosθ - cosθ + θ sinθ)

= aθsinθ - (ii)

Dividing (ii) by (i) -

\(\frac{dy}{dx}\) = tanθ 

⇒ \(\frac{dy}{dx}\) at (\(\theta = \frac{\pi }{4}\)) = 1

Calculation:

Given: y = a cos 2x + b sin 2x

Now, by differentiating both the sides w.r.t x, we get

\(\Rightarrow \frac{{dy}}{{dx}} = \; - 2a\sin 2x + 2b\cos 2x\)

Now, again by differentiating both the sides w.r.t x of the above equation, we get

\(\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \; - \;4a\cos 2x - 4b\sin 2x\)

\(\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \; - 4\;\left( {a\cos 2x + b\sin 2x} \right) = \; - \;4y\)

\(\therefore The \space \space required \space \space differential \space equation \space \space is \space \frac{{{d^2}y}}{{d{x^2}}} + 4y = 0\)

Concept:

If a differential equation contains n arbitrary constants, then differentiate the equation n times.

Calculation:

Given differential equation is y = ax2 + bx + c

Here a, b, c are the arbitrary constants.

Differentiating both sides with respect to x,

\(\rm\frac{dy}{dx}=2ax +b\)

Again differentiating, we get:

\(\rm\frac{d^2y}{dx^2}=2a\)

Again differentiating, we have:

\(\rm\frac{d^3y}{dx^3}=0\)

The required differential equation is \(\rm\frac{d^3y}{dx^3}=0\). 

Concept:

Equation of parabola with y-axis and vertex (0, k) is 

(x - 0)² = 4a (y - k).

Calculations:

Given, equation of parabola with y-axis and vertex (0, k) is 

(x - 0)² = 4a (y - k).

⇒ x² = 4ay - 4ak 

Taking derivative on both side, we get.

\(\)⇒\(\rm 2x = 4a \dfrac{dy}{dx}\) 

⇒\(\rm \dfrac{1}{x}\dfrac{dy}{dx}= \dfrac{1}{2a}\)

Again taking derivative on both side, we get.

⇒ \(\rm \dfrac{d}{dx}(\dfrac{1}{x}\dfrac{dy}{dx})= \dfrac{d}{dx}(\dfrac{1}{2a})\)

⇒\(\rm \dfrac{1}{x}\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}(\dfrac{-1}{x^2})= 0\)

⇒\(\rm x.\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx} = 0\)

Hence, the differential equation of all parabolas whose axis is y-axis is \(\rm x.\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx} = 0\)

Concept:

Some useful formulas are:

\(\rm{ d(e^{ax})\over dx} = ae^{ax}\)

Calculation:

Given function is,

y = mex + ne^-x

Differentiating the function we get,

\(\rm{ dy\over dx} = me^x-ne^{-x}\)

Differentiating further with respect to x, we get

\(\rm{ d^2y\over dx^2} = me^x+ne^{-x}\)

Differentiating further with respect to x, we get

\(\rm{ d^3y\over dx^3} = me^x-ne^{-x}\)

Now, \(\rm{ d^3y\over dx^3}+{ d^2y\over dx^2}-{ dy\over dx}-y \)

= \(\rm me^x-ne^{-x}+me^x+ne^{-x}-me^x+ne^{-x}-me^x-ne^{-x}\)

= 0

Hence, \(\rm{ d^3y\over dx^3}+{ d^2y\over dx^2}-{ dy\over dx}-y =0\)

Formula Used:

d(cosx) = - sinx

d(sinx) = cosx

Calculation:

y = 4cos3x

Differentiate with respect to x

dy/dx = 4 × (- 3sin3x)

Again differential with respect to x

⇒ \(\frac{d^2 y}{d x^2}\) = -12 × (3cos3x)

⇒ \(\frac{d^2 y}{d x^2}\) = - 9 × (4cos3x)

⇒ \(\frac{d^2 y}{d x^2}\) = - 9y

∴​ \(\frac{d^2 y}{d x^2}\) + 9y = 0