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

7 Marks

LAQ-1 : Solve (x2+y2)dx=2xydy

Given D.E is $$(x^2 + y^2)dx = 2xydy$$

$$\Rightarrow \frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$$

Since (1) is a homogeneous D.E,

Let y = vx

$$\Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}$$

Substituting into (1) gives:

$$v + x \frac{dv}{dx} = \frac{x^2 + (vx)^2}{2x(vx)} = \frac{x^2 + v^2x^2}{2x^2v} = \frac{1 + v^2}{2v}$$

$$x \frac{dv}{dx} = \frac{1 + v^2}{2v} – v = \frac{1 + v^2 – 2v^2}{2v} = \frac{1 – v^2}{2v}$$

$$\Rightarrow 2v \frac{dv}{1 – v^2} = \frac{dx}{x}$$

$$\Rightarrow \int 2v \frac{dv}{1 – v^2} = \int \frac{dx}{x}$$

$$\Rightarrow -\log|1 – v^2| = \log|x| + \log|c|$$

$$\Rightarrow \log|x| + \log|1 – v^2| = \log|c|$$

$$\Rightarrow \log|x(1 – v^2)| = \log|c|$$

$$\Rightarrow x(1 – v^2) = c$$

$$\Rightarrow x(1 – \frac{y^2}{x^2}) = c$$

$$\Rightarrow x^2 – y^2 = cx$$


LAQ-2 : Solve (x2-y2)dx-xydy=0

Given D.E is $$(x^2 – y^2)dx – xydy = 0$$

$$\Rightarrow \frac{dy}{dx} = \frac{x^2 – y^2}{xy}$$

Since (1) is a homogeneous D.E,

Let y = vx

$$\Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}$$

Substituting into (1),

$$\Rightarrow v + x \frac{dv}{dx} = \frac{x^2 – v^2x^2}{xv} = \frac{x^2(1 – v^2)}{x^2 v} = \frac{1 – v^2}{v}$$

$$\Rightarrow x \frac{dv}{dx} = \frac{1 – v^2}{v} – v = \frac{1 – v^2 – v^2}{v} = \frac{1 – 2v^2}{v}$$

$$\Rightarrow \frac{v}{1 – 2v^2} dv = \frac{dx}{x}$$

$$\Rightarrow \int \frac{v}{1 – 2v^2} dv = \int \frac{dx}{x}$$

$$\Rightarrow -\frac{1}{4} \log |1 – 2v^2| = \log x + \log c$$

$$\Rightarrow \log(1 – 2v^2) = -4 \log cx$$

$$\Rightarrow 1 – 2v^2 = \frac{1}{cx^4}$$

$$\Rightarrow 1 – 2\frac{y^2}{x^2} = \frac{1}{cx^4}$$

$$\Rightarrow 1 – 2\frac{y^2}{x^2} = \frac{1}{cx^4}$$

$$\Rightarrow x^2 – 2y^2 = \frac{1}{cx^2}$$

$$\Rightarrow cx^2(x^2 – 2y^2) = 1$$

$$\Rightarrow x^4 – 2x^2y^2 = c$$


LAQ-3 : Find the equation of a curve whose gradient is dy/dx=y/x-cos2 (y/x), which passes through the point (1,π/4)

Given D.E is $$\frac{dy}{dx} = \frac{y}{x} – \cos^2\left(\frac{y}{x}\right)$$

Since (1) is a homogeneous D.E,

Let y = vx then $$\frac{dy}{dx} = v + x \frac{dv}{dx}$$

$$\Rightarrow v + x \frac{dv}{dx} = \frac{vx}{x} – \cos^2\left(\frac{vx}{x}\right) = v – \cos^2(v)$$

$$\Rightarrow x \frac{dv}{dx} = v – \cos^2(v) – v$$

$$\Rightarrow x \frac{dv}{dx} = -\cos^2(v)$$

$$\Rightarrow \frac{dv}{\cos^2(v)} = -\frac{dx}{x}$$

$$\Rightarrow \int \sec^2(v) dv = -\int \frac{dx}{x}$$

$$\Rightarrow \tan(v) = -\log|x| + c$$

Letting $$v = \frac{y}{x}$$

$$\Rightarrow \tan\left(\frac{y}{x}\right) + \log|x| = c$$

If this curve passes through the point $$(x, y) = (1, \frac{\pi}{4})$$ then

$$\tan\left(\frac{\pi}{4}\right) + \log(1) = c$$

$$\Rightarrow 1 + 0 = c$$

$$\Rightarrow c = 1$$

Thus, from (2) the required solution is

$$\tan\left(\frac{y}{x}\right) + \log|x| = 1$$


LAQ-4 : Give solution of xsiny/x dx = ydx-xdy which passes through the point (1,π∕4)

The given D.E is written as

$$xdy = ydx – x \sin^2\left(\frac{y}{x}\right)dx$$

$$\Rightarrow xdy = (y – x \sin^2\left(\frac{y}{x}\right))dx$$

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

Since (1) is a homogeneous D.E,

Put y = vx

$$\Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}$$

$$\Rightarrow v + x\frac{dv}{dx} = \frac{vx}{x} – \sin^2\left(\frac{vx}{x}\right) = v – \sin^2(v)$$

$$\Rightarrow x\frac{dv}{dx} = v – \sin^2(v) – v$$

$$\Rightarrow x\frac{dv}{dx} = -\sin^2(v)$$

$$\Rightarrow -\frac{dv}{\sin^2(v)} = \frac{dx}{x}$$

$$\Rightarrow \int -\frac{dv}{\sin^2(v)} = \int \frac{dx}{x}$$

$$\Rightarrow \int -\csc^2(v)dv = \int \frac{1}{x}dx$$

$$\Rightarrow \cot(v) = \log(x) + c$$

$$\Rightarrow \cot\left(\frac{y}{x}\right) = \log(x) + c$$

If this curve passes through the point

$$(x, y) = (1, \frac{\pi}{4})$$ then

$$\cot\left(\frac{\pi}{4}\right) = \log(1) + c$$

$$\Rightarrow 1 = 0 + c$$

$$\Rightarrow c = 1$$

From (2), the required solution is

$$\cot\left(\frac{y}{x}\right) = \log(x) + 1$$


LAQ-5 : Solve (x2y – 2xy2)dx = (x3 – 3x2y)dy

Given D.E is $$(x^2y – 2xy^2)dx = (x^3 – 3x^2y)dy$$

$$\Rightarrow \frac{dy}{dx} = \frac{x^2y – 2xy^2}{x^3 – 3x^2y}$$

Since (1) is a homogeneous D.E,

Let y = vx

$$\Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}$$

$$\Rightarrow v + x\frac{dv}{dx} = \frac{x^2(vx) – 2x(v^2x^2)}{x^3 – 3x^2(vx)} = \frac{vx^3 – 2v^2x^3}{x^3 – 3vx^3} = \frac{v – 2v^2}{1 – 3v}$$

$$\Rightarrow x\frac{dv}{dx} = \frac{v – 2v^2}{1 – 3v} – v = \frac{v – 2v^2 – v + 3v^2}{1 – 3v} = \frac{v^2}{1 – 3v}$$

$$\Rightarrow \frac{dx}{x} = \frac{1 – 3v}{v^2} dv$$

$$\Rightarrow \int \frac{dx}{x} = \int \left(\frac{1}{v^2} – 3\frac{1}{v}\right) dv$$

$$\Rightarrow \log x + \log c = -\frac{1}{v} – 3\log v$$

$$\Rightarrow \log(x) + \log c + 3\log v = -\frac{1}{v}$$

$$\Rightarrow \log(x) + \log c + \log(v^3) = -\frac{1}{v}$$

$$\Rightarrow \log(xv^3c) = -\frac{1}{v}$$

$$\Rightarrow xv^3c = e^{-1/v}$$

$$\Rightarrow xv^3c = e^{-1/v}$$

$$\Rightarrow x(y^3/x^3)c = e^{-x/y}$$

$$\Rightarrow c = \frac{x^2}{y^3} e^{-x/y}$$


LAQ-6 : Solve (x3 – 3xy2)dx+(3x2y-y3)dy=0

Given D.E is

$$(x^3 – 3xy^2)dx + (3x^2y – y^3)dy = 0$$

$$\Rightarrow (x^3 – 3xy^2)dx = (y^3 – 3x^2y)dy$$

$$\Rightarrow \frac{dy}{dx} = \frac{x^3 – 3xy^2}{y^3 – 3x^2y}$$

Since (1) is a homogeneous D.E,

Let y = vx

$$\Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}$$

$$\Rightarrow v + x \frac{dv}{dx} = \frac{x^3(1 – 3v^2)}{x^3(v^3 – 3v)} = \frac{1 – 3v^2}{v^3 – 3v}$$

$$\Rightarrow x \frac{dv}{dx} = \frac{1 – 3v^2}{v^3 – 3v} – v$$

$$\Rightarrow x \frac{dv}{dx} = \frac{1 – 3v^2 – v^4 + 3v^2}{v^3 – 3v}$$

$$\Rightarrow x \frac{dv}{dx} = \frac{v^4 – 1}{v^3 – 3v}$$

$$\Rightarrow \frac{3v – v^3}{v^4 – 1} dv = \frac{dx}{x}$$

Writing the L.H.S as partial fractions,

$$\int \left[\frac{1}{2}\left(\frac{1}{v + 1}\right) + \frac{1}{2}\left(\frac{1}{v – 1}\right) – \frac{2v}{v^2 + 1}\right]dv = \int \frac{dx}{x}$$

$$\Rightarrow \frac{1}{2} \log|v + 1| + \frac{1}{2} \log|v – 1| – \log|v^2 + 1| = \log|x| + \log|c|$$

$$\Rightarrow \log\left[\sqrt{|v + 1|} \cdot \sqrt{|v – 1|}\right] – \log|v^2 + 1| = \log|cx|$$

$$\Rightarrow \log\left[\frac{\sqrt{|v^2 – 1|}}{|v^2 + 1|}\right] = \log|cx|$$

$$\Rightarrow \frac{\sqrt{|v^2 – 1|}}{|v^2 + 1|} = cx$$

$$\Rightarrow \frac{\sqrt{|y^2/x^2 – 1|}}{|y^2/x^2 + 1|} = cx$$

$$\Rightarrow \frac{\sqrt{y^2 – x^2}}{y^2 + x^2} = cx$$

$$\Rightarrow \frac{y^2 – x^2}{y^2 + x^2} = c^2x^2$$

$$\Rightarrow y^2 – x^2 = c^2x^2(y^2 + x^2)$$


LAQ-7 : Solve (y2-2xy)dx+(2xy-x2)dy=0

Given D.E is $$(y^2 – 2xy)dx + (2xy – x^2)dy = 0$$

$$\Rightarrow \frac{dy}{dx} = \frac{y^2 – 2xy}{x^2 – 2xy}$$

This is a homogeneous D.E put

$$y = vx$$

$$\Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}$$

From (1)

$$\Rightarrow v + x \frac{dv}{dx} = \frac{v^2x^2 – 2x^2v}{x^2 – 2x^2v} = \frac{x^2(v^2 – 2v)}{x^2(1 – 2v)} = \frac{v^2 – 2v}{1 – 2v}$$

$$\Rightarrow x \frac{dv}{dx} = \frac{v^2 – 2v}{1 – 2v} – v = \frac{v^2 – 2v – v(1 – 2v)}{1 – 2v} = \frac{3v^2 – 3v}{-(2v – 1)}$$

$$\Rightarrow x \frac{dv}{dx} = \frac{3(v^2 – v)}{-(2v – 1)}$$

$$\Rightarrow \frac{(2v – 1)}{v^2 – v} dv = -3 \frac{dx}{x}$$

$$\Rightarrow \int \frac{(2v – 1)}{v^2 – v} dv = -3 \int \frac{dx}{x}$$

$$\Rightarrow \log|v^2 – v| = -3 \log|x| + \log|c|$$

$$\Rightarrow \log|v^2 – v| = \log|x^{-3}| + \log|c|$$

$$\Rightarrow \log|v^2 – v| = \log|c/x^3|$$

$$\Rightarrow v^2 – v = \frac{c}{x^3}$$

$$\Rightarrow y^2/x^2 – y/x = \frac{c}{x^3}$$

$$\Rightarrow y^2 – xy = cx^{-1}$$

$$\Rightarrow xy(y – x) = c$$


LAQ-8 : Solve dy/dx=y2-2xy/x2-xy

Given D.E is $$\frac{dy}{dx} = \frac{y^2 – 2xy}{x^2 – xy}$$

This is a homogeneous D.E.

Put y = vx

$$\Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}$$

From (1),

$$\Rightarrow v + x \frac{dv}{dx} = \frac{v^2x^2 – 2x^2v}{x^2 – x^2v} = \frac{x^2(v^2 – 2v)}{x^2(1 – v)} = \frac{v^2 – 2v}{1 – v}$$

$$\Rightarrow x \frac{dv}{dx} = \frac{v^2 – 2v}{1 – v} – v = \frac{v^2 – 2v – v + v^2}{1 – v} = \frac{2v^2 – 3v}{1 – v}$$

$$\Rightarrow x \frac{dv}{dx} = \frac{2v^2 – 3v}{1 – v}$$

$$\Rightarrow \frac{1 – v}{2v^2 – 3v} dv = \frac{dx}{x}$$

$$\Rightarrow \int \frac{1 – v}{2v^2 – 3v} dv = \int \frac{dx}{x}$$

$$\Rightarrow \int \frac{1 – v}{v(2v – 3)} dv = \log x + \log c$$

$$\Rightarrow – \frac{1}{3} \left[\log v + \frac{1}{2} \log(2v – 3)\right] = \log x + \log c$$

$$\Rightarrow \log v\sqrt{2v – 3} = -3 \log x/c$$

$$\Rightarrow v\sqrt{2v – 3} = \frac{c^3}{x^3}$$

$$\Rightarrow x^3v\sqrt{2v – 3} = c^3$$

$$\Rightarrow x^6v^2(2v – 3) = c^6$$

$$\Rightarrow x^6\left(\frac{y^2}{x^2}\right)\left(2 \frac{y}{x} – 3\right) = c$$

$$\Rightarrow x^3y^2(2y – 3x) = c$$


LAQ-9 : Solve the differential equation (2x + y + 1)dx + (4x + 2y – 1) dy = 0

Given that $$(4x + 2y – 1)dy = -(2x + y + 1)dx$$

$$\Rightarrow \frac{dy}{dx} = -\frac{2x + y + 1}{4x + 2y – 1}$$

Comparing 2x + y + 1 and 4x + 2y – 1 with

$$\frac{a_1x + b_1y + c_1}{a_2x + b_2y + c_2}$$ we get

$$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}; \quad \frac{b_1}{b_2} = \frac{1}{2}$$

$$\Rightarrow \frac{a_1}{a_2} = \frac{b_1}{b_2}$$

So we take the substitution

$$2x + y = v$$

$$\Rightarrow 2 + \frac{dy}{dx} = \frac{dv}{dx}$$

$$\Rightarrow \frac{dy}{dx} = \frac{dv}{dx} – 2$$

Substitute $$\frac{dy}{dx}$$ in 1

$$\frac{dv}{dx} – 2 = -\frac{v + 1}{2v – 1}$$

$$\Rightarrow \frac{dv}{dx} = 2 – \frac{v + 1}{2v – 1}$$

$$\Rightarrow \frac{dv}{dx} = \frac{4v – 2 – v – 1}{2v – 1} = \frac{3v – 3}{2v – 1} = 3\frac{v – 1}{2v – 1}$$

$$\Rightarrow \frac{dx}{dv} = \frac{1}{3}\frac{2v – 1}{v – 1}$$

$$\Rightarrow dx = \frac{1}{3}\left(2 + \frac{1}{v – 1}\right)dv$$

Integrating both sides we get

$$\int dx = \frac{1}{3} \int 2dv + \frac{1}{3} \int \frac{dv}{v – 1}$$

$$\Rightarrow x = \frac{2}{3}v + \frac{1}{3}\log|v – 1| + C$$

$$\Rightarrow 3x = 2(2x + y) + \log|2x + y – 1| + C$$

$$\Rightarrow 3x = 4x + 2y + \log|2x + y – 1| + C$$

$$\Rightarrow x + 2y + \log|2x + y – 1| + C = 0$$


LAQ-10 : Solve dy/dx= x-y+3/2x-2y+5

Given that

$$\frac{dy}{dx} = \frac{x – y + 3}{2x – 2y + 5} = \frac{(x – y) + 3}{2(x – y) + 5}$$

Comparing x – y + 3 and 2x – 2y + 5 with

$$\frac{a_1x + b_1y + c_1}{a_2x + b_2y + c_2}$$ we get

$$\frac{a_1}{a_2} = 1/2; \quad \frac{b_1}{b_2} = -1/-2 = 1/2$$

$$\Rightarrow \frac{a_1}{a_2} = \frac{b_1}{b_2}$$

So we take the substitution

$$x – y = v$$

$$\Rightarrow 1 – \frac{dy}{dx} = \frac{dv}{dx}$$

$$\Rightarrow \frac{dy}{dx} = 1 – \frac{dv}{dx}$$

$$\Rightarrow 1 – \frac{dv}{dx} = v + 3/2v + 5$$

$$\Rightarrow \frac{dv}{dx} = 1 – v – 3/2v – 5$$

$$\Rightarrow \frac{dv}{dx} = -v – 2/2v – 5$$

$$\Rightarrow \frac{dx}{dv} = 2v + 5/v + 2$$

$$\Rightarrow dx = \left(2v + 5\right) / \left(v + 2\right) dv = \left(2v + 4 + 1\right) / \left(v + 2\right) dv$$

$$= \left(2(v + 2) + 1\right) / \left(v + 2\right) dv = \left(2 + \frac{1}{v + 2}\right) dv$$

Integrating on both sides we get

$$\int dx = \int 2dv + \int \frac{dv}{v + 2}$$

$$\Rightarrow x = 2v + \log|v + 2| + c$$

$$\Rightarrow x = 2(x – y) + \log|(x – y) + 2| + c$$

$$\Rightarrow x – 2(x – y) + \log(x – y + 2) = c$$

$$\Rightarrow x – 2y + \log(x – y + 2) = c$$