15 Most VSAQ’s of Differential Equations Chapter in Inter 2nd Year Maths-2B (TS/AP)

2 Marks

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

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VSAQ-2 : Find the order and degree of the D.E x^1∕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

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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

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VSAQ-4 : Form the differential equation corresponding to y = A cos 3x + B sin 3x,where A and B are parameters

Given $$y = A \cos 3x + B \sin 3x$$

$$\Rightarrow \frac{dy}{dx} = -3A(\sin 3x) + 3B(\cos 3x)$$

$$\Rightarrow \frac{d^2y}{dx^2} = -9A \cos 3x – 9B \sin 3x = -9(A \cos 3x + B \sin 3x) = -9y$$

$$\frac{d^2y}{dx^2} + 9y = 0$$

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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) $$y = y’x – 2y’^2$$

$$\Rightarrow 2y’^2 – xy’ + y = 0$$

$$\Rightarrow 2\left(\frac{dy}{dx}\right)^2 – x\left(\frac{dy}{dx}\right) + y = 0$$

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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 $$2x + 2y \frac{dy}{dx} = 0$$

$$\Rightarrow x + y \frac{dy}{dx} = 0$$

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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 $$y” = -an^2\cos(nx + b)$$

$$\Rightarrow y” = -n^2a\cos(nx + b)$$

$$\Rightarrow y” + n^2y = 0$$

$$\Rightarrow y” + n^2y = 0$$

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VSAQ-8 : Find the general solution of dy/dx = e^x+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 $$\int e^x dx – \int e^{-y} dy = c$$

$$\Rightarrow e^x + e^{-y} = c$$

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VSAQ-9 : Solve dy/dx + 1 = e^x+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) $$\Rightarrow \frac{dt}{dx} = e^t$$

$$\Rightarrow \frac{dt}{e^t} = dx$$

$$\Rightarrow \int \frac{dt}{e^t} = \int dx$$

$$\Rightarrow -e^{-t} = x + c$$

$$\Rightarrow e^{-(x + y)} + x + c = 0$$

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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 $$\tan^{-1}y = \tan^{-1}x + c$$

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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 $$\sin^{-1}y + \sin^{-1}x = c$$

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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 $$y^2 + x^2 = 2c$$

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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 $$y = c x^2$$

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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) = $$e^{\int P(x)dx} = e^{\int \tan x dx} = e^{\log \sec x} = \sec x$$

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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) = $$e^{\int P(x)dx} = e^{\int -\frac{1}{x} dx} = e^{-\log x} = e^{\log \frac{1}{x}} = \frac{1}{x}$$