11 Most SAQ’s of Differential Equations Chapter in Inter 2nd Year Maths-2B (TS/AP)
4 Marks
SAQ-1 : Solve (xy2+x)dx+(yx2+y)dy=0
The given D.E is in the variables and separable form
$$(xy^2 + x)dx + (yx^2 + y)dy = 0$$
$$\Rightarrow (xy^2 + x)dx = -(yx^2 + y)dy$$
$$\Rightarrow x(y^2 + 1)dx = -y(x^2 + 1)dy$$
$$\Rightarrow \frac{x}{x^2 + 1} dx = – \frac{y}{y^2 + 1} dy$$
$$\Rightarrow \int \frac{2x}{x^2 + 1} dx = – \int \frac{2y}{y^2 + 1} dy$$
$$\Rightarrow \log(x^2 + 1) = -\log(y^2 + 1) + \log c$$
$$\Rightarrow \log(x^2 + 1) + \log(y^2 + 1) = \log c$$
$$\Rightarrow \log((x^2 + 1)(y^2 + 1)) = \log c$$
$$\Rightarrow (x^2 + 1)(y^2 + 1) = c$$
The solution is $$(x^2 + 1)(y^2 + 1) = c$$
SAQ-2 : Solve dy/dx – xtan(y-x) = 1
Put $$y – x = t$$
$$\Rightarrow \frac{dy}{dx} – 1 = \frac{dt}{dx}$$
$$\Rightarrow \frac{dy}{dx} = \frac{dt}{dx} + 1$$
Hence, the given D.E becomes $$(\frac{dt}{dx} + 1) – x \tan t = 1$$
$$\Rightarrow \frac{dt}{dx} = x \tan t$$
$$\Rightarrow \frac{1}{\tan t} dt = x dx$$
$$\Rightarrow \int \cot t dt = \int x dx$$
$$\Rightarrow \log(\sin t) = \frac{x^2}{2} + C$$
$$\Rightarrow 2 \log(\sin t) = x^2 + C$$
$$\Rightarrow 2 \log(\sin(y – x)) = x^2 + C$$
SAQ-3 : Solve dy/dx + y tanx = cos3 x
Given D.E is $$\frac{dy}{dx} + y \tan x = \cos^3 x$$
The equation is in the form $$\frac{dy}{dx} + P(x)y = Q(x)$$ which is a linear D.E in y.
Here $$P = \tan x$$
$$\Rightarrow \int Pdx = \int \tan x dx = \log(\sec x)$$
The Integrating Factor (I.F) is $$e^{\int Pdx} = e^{\log(\sec x)} = \sec x$$
Hence, the solution is $$y \cdot \text{I.F} = \int (\text{I.F} \cdot Q)dx$$
$$\Rightarrow y \cdot \sec x = \int \sec x \cdot \cos^3 x dx = \int \cos^3 x \cdot \frac{1}{\cos x} dx = \int \cos^2 x dx$$
$$= \frac{1}{2} \int (1 + \cos 2x)dx = \frac{1}{2}(x + \frac{\sin 2x}{2}) + C$$
$$\Rightarrow y \cdot \sec x = \frac{1}{2}x + \frac{1}{4}\sin 2x + C$$
$$\Rightarrow 2y \cdot \sec x = x + \frac{1}{2}\sin 2x + 2C$$
$$\Rightarrow 2y = x \cos x + \sin x \cos x + 2C \cdot \cos x$$
The solution is $$2y = x \cos x + \sin x \cos x + C \cdot \cos x$$
SAQ-4 : Solve dy/dx + y tanx = sinx
Given D.E is in the form $$\frac{dy}{dx} + yP(x) = Q(x)$$
This is a linear D.E in y.
Here $$P = \tan x$$
$$\Rightarrow \int Pdx = \int \tan x dx = \log(\sec x)$$
The Integrating Factor (I.F) is $$e^{\int Pdx} = e^{\log(\sec x)} = \sec x$$
Hence, the solution is $$y \cdot \text{I.F} = \int (\text{I.F} \cdot Q)dx$$
$$\Rightarrow y \cdot \sec x = \int \sec x \cdot \sin x dx = \int \tan x dx = \log(\sec x) + C$$
The solution is $$y \cdot \sec x = \log(\sec x) + C$$
SAQ-5 : Solve cosx.dy/dx + ysinx = sec2 x
Given D.E is $$\cos x \cdot \frac{dy}{dx} + y \sin x = \sec^2 x$$
$$\Rightarrow \frac{dy}{dx} + y \left( \frac{\sin x}{\cos x} \right) = \frac{\sec^2 x}{\cos x}$$
$$\Rightarrow \frac{dy}{dx} + y (\tan x) = \sec^3 x$$
The given D.E is in the form $$\frac{dy}{dx} + P(x)y = Q(x)$$
This is a linear D.E in y.
Here $$P = \tan x$$
$$\Rightarrow \int Pdx = \int \tan x dx = \log(\sec x)$$
The Integrating Factor (I.F) is $$e^{\int Pdx} = e^{\log(\sec x)} = \sec x$$
Hence, the solution is $$y \cdot \text{I.F} = \int (\text{I.F} \cdot Q) dx$$
$$\Rightarrow y \cdot \sec x = \int \sec x \cdot \sec^3 x dx = \int \sec^4 x dx$$
$$= \int (1 + \tan^2 x) \sec^2 x dx = \int \sec^2 x dx + \int \sec^2 x \tan^2 x dx$$
$$= \tan x + \frac{\tan^3 x}{3} + C$$
$$\Rightarrow y = \left( \tan x + \frac{\tan^3 x}{3} + C \right) \cdot \cos x$$
SAQ-6 : Solve dy/dx + y secx = tanx
Given D.E is in the form $$\frac{dy}{dx} + yP(x) = Q(x)$$
This is a linear D.E in y.
Here $$P = \sec x$$
$$\Rightarrow \int \sec x dx = \log | \sec x + \tan x |$$
The Integrating Factor (I.F) is $$e^{\int Pdx} = e^{\log|\sec x + \tan x|} = \sec x + \tan x$$
Hence, the solution is $$y \cdot \text{I.F} = \int (\text{I.F} \cdot Q)dx$$
$$\Rightarrow y(\sec x + \tan x) = \int (\sec x + \tan x) \tan x dx = \int (\sec x \tan x + \tan^2 x)dx$$
$$= \int (\sec x \tan x + \sec^2 x – 1)dx = \int \sec x \tan x dx + \int \sec^2 x dx – \int dx$$
$$= \sec x + \tan x – x + C$$
The solution is $$y(\sec x + \tan x) = \sec x + \tan x – x + C$$
SAQ-7 : Solve (1 + x2)dy/dx+y = eTan-1 x
Given D.E is $$(1 + x^2)\frac{dy}{dx} + y = e^{\tan^{-1}x}$$
$$\Rightarrow \frac{dy}{dx} + y\left(\frac{1}{1 + x^2}\right) = \frac{e^{\tan^{-1}x}}{1 + x^2}$$
The above equation is in the form $$\frac{dy}{dx} + yP(x) = Q(x)$$
This is a linear D.E in y.
Here $$P = \frac{1}{1 + x^2}$$
$$\Rightarrow \int Pdx = \int \frac{1}{1 + x^2} dx = \tan^{-1} x$$
$$e^{\int Pdx} = e^{\tan^{-1}x}$$
Hence, the solution is $$y \cdot \text{I.F} = \int (\text{I.F} \cdot Q) dx$$
$$\Rightarrow y \cdot e^{\tan^{-1}x} = \int e^{\tan^{-1}x} \cdot \frac{e^{\tan^{-1}x}}{1 + x^2} dx$$
Put $$\tan^{-1}x = t$$
$$\Rightarrow \frac{1}{1 + x^2} dx = dt$$
$$y \cdot e^t = \int e^t \cdot e^t dt = \int e^{2t} dt = \frac{1}{2} e^{2t} + C$$
$$\Rightarrow y \cdot e^{\tan^{-1}x} = \frac{1}{2} e^{2\tan^{-1}x} + C$$
SAQ-8 : Solve dy/dx + 2y/x = 2x2
Given D.E is $$\frac{dy}{dx} + y\left(\frac{2}{x}\right) = 2x^2$$
This is a linear D.E in y.
It is in the form $$\frac{dy}{dx} + yP(x) = Q(x)$$ where $$P(x) = \frac{2}{x}$$ and $$Q(x) = 2x^2$$
Now I.F = $$e^{\int P(x)dx} = e^{\int \frac{2}{x} dx} = e^{2\log x} = e^{\log x^2} = x^2$$
The solution is $$y \cdot \text{I.F} = \int (\text{I.F} \cdot Q(x))dx + C$$
$$\Rightarrow y \cdot x^2 = \int x^2 \cdot 2x^2 dx = 2 \int x^4 dx = \frac{2}{5}x^5 + C$$
$$\Rightarrow yx^2 = \frac{2}{5}x^5 + C$$
$$\Rightarrow 5yx^2 = 2x^5 + C$$
SAQ-9 : Solve (1+y2)dx = (Tan-1 y-x)dy
Given D.E is $$(1 + y^2)dx = (\tan^{-1} y – x)dy$$
$$\Rightarrow \frac{dx}{dy} = \frac{\tan^{-1}y}{1 + y^2}$$
$$\Rightarrow \frac{dx}{dy} + x\left(\frac{1}{1 + y^2}\right) = \frac{\tan^{-1}y}{1 + y^2}$$
The above equation is in the form $$\frac{dx}{dy} + xP(y) = Q(y)$$ which is a linear D.E in x.
Here $$P = \frac{1}{1 + y^2}$$
$$\Rightarrow \int Pdy = \int \frac{1}{1 + y^2} dy = \tan^{-1}y$$
$$e^{\int Pdy} = e^{\tan^{-1}y}$$
The solution is $$x \cdot \text{I.F} = \int (\text{I.F} \cdot Q)dy$$
$$\Rightarrow x \cdot e^{\tan^{-1}y} = \int e^{\tan^{-1}y}\left(\frac{\tan^{-1}y}{1 + y^2}\right)dy$$
Let $$\tan^{-1}y = t$$ then $$1/(1 + y^2) dy = dt$$
$$\Rightarrow x \cdot e^t = \int e^t t dt = e^t(t – 1) + C$$ where $$e^t(t – 1) + C$$ is the integration of $$e^t t$$
$$\Rightarrow x \cdot e^{\tan^{-1}y} = (tan^{-1}y – 1)e^{\tan^{-1}y} + C$$
$$\Rightarrow x = \tan^{-1}y – 1 + C e^{-\tan^{-1}y}$$
SAQ-10 : Solve (x + y + 1)dy/dx = 1
Given D.E is $$(x + y + 1)\frac{dy}{dx} = x + y + 1$$
$$\Rightarrow \frac{dx}{dy} – x = y + 1$$
The above equation is in the form $$\frac{dx}{dy} + xP(y) = Q(y)$$ which is a linear D.E in x.
Here $$P = -1$$
$$\Rightarrow \text{I.F} = e^{\int Pdy} = e^{\int -1 dy} = e^{-y}$$
Hence, the solution is $$x \cdot \text{I.F} = \int (\text{I.F} \cdot Q)dy$$
$$\Rightarrow x \cdot e^{-y} = \int e^{-y}(y + 1)dy$$
To integrate $$\int e^{-y}(y + 1)dy$$ perform integration by parts
$$\Rightarrow \int e^{-y}(y + 1)dy = \int y e^{-y} dy + \int e^{-y} dy$$
$$\Rightarrow -(y + 1)e^{-y} – e^{-y} + C = -(y + 2)e^{-y} + C$$
$$\Rightarrow x \cdot e^{-y} = -(y + 2)e^{-y} + C$$
$$\Rightarrow x + y + 2 = Ce^y$$
SAQ-11 : Solve xlogx dy/dx + y = 2logx
Given D.E is $$x \log x \frac{dy}{dx} + y = 2 \log x$$
$$\Rightarrow \frac{dy}{dx} + y\left(\frac{1}{x \log x}\right) = \frac{2}{x}$$
The above equation is in the form $$\frac{dy}{dx} + yP(x) = Q(x)$$
which is a linear D.E in y
Here $$P(x) = \frac{1}{x \log x}$$
$$\Rightarrow \int P(x)dx = \int \frac{1}{x \log x} dx = \log(\log x)$$
$$e^{\int P(x)dx} = e^{\log(\log x)} = \log x$$
The solution is $$y \cdot \text{I.F} = \int (\text{I.F} \cdot Q(x))dx$$
$$\Rightarrow y(\log x) = \int \log x \cdot \frac{2}{x} dx = 2 \int \log x \frac{dx}{x}$$
Put $$\log x = t$$
$$\Rightarrow \frac{1}{x} dx = dt$$
$$y \cdot \log x = 2 \int t dt = 2 \left(\frac{t^2}{2}\right) + C$$
$$\Rightarrow y \cdot \log x = t^2 + C$$
$$\Rightarrow y(\log x) = (\log x)^2 + C$$