Solving Homogeneous Differential Equation MCQ Quiz in தமிழ் - Objective Question with Answer for Solving Homogeneous Differential Equation - இலவச PDF ஐப் பதிவிறக்கவும்

Calculation

\((x^2-y^2)dx+2xydy=0\)

\(\Rightarrow \dfrac{dy}{dx}=\dfrac{y^2-x^2}{2xy}\)

Put \(y=vx\)  ⇒ \(\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}\)

\(\Rightarrow v+x\dfrac{dv}{dx}=\dfrac{v^2x^2-x^2}{2vx^2}\)

\(\Rightarrow v+x\dfrac{dv}{dx}=\dfrac{v^2-1}{2v}\)

\(\Rightarrow x\dfrac{dv}{dx}=\dfrac{-v^2-1}{2v}\)

\(\Rightarrow \dfrac{2vdv}{v^2+1}=-\dfrac{dx}{x}\)

Integrating we get;

⇒ \(ln\vert v^2+1\vert=-ln\vert x\vert+lnc\)

⇒ \(\dfrac{y^2}{x^2}+1=\dfrac{c}{x}\)

Putting (1,1)

⇒ \(c=2\)

⇒ \(x^2+y^2-2x=0\)

hence its is a circle of radius 1

Hence option 2 is correct

Calculation

Given

f(1) = 1,  f(a) = 0 

f(x) = \(\left.\sqrt{\lim _{x \rightarrow x}\left\{\left.\frac{2 r^2\left[(f(r))^2-f(x) f(r)\right]}{r^2-x^2}-r^3 e^{\frac{f(r)}{r}} \right\rvert\,\right.}\right\}\)

⇒ \(f^2(x)=\operatorname{Lim}_{r \rightarrow x}\left(\frac{2 r^2\left(f^2(r)-f(x) f(r)\right)}{r^2-x^2}-r^3 e^{\frac{f(r)}{r}}\right)\)

⇒  \(\operatorname{Lim}_{r \rightarrow x}\left(\frac{2 r^2 f(r)}{r+x} \frac{(f(r)-f(x))}{r-x}-r^3 e^{\frac{f(r)}{r}}\right)\)

⇒ \(f^2(x)=\frac{2 x^2 f(x)}{2 x} f^{\prime}(x)-x^3 e^{\frac{f(x)}{x}}\)

⇒ \(y^2=x y \frac{d y}{d x}-x^3 e^{\frac{y}{x}}\)

⇒ \(\frac{y}{x}=\frac{d y}{d x}-\frac{x^2}{y} e^{\frac{y}{x}}\)

Put \(y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)

⇒ \(v=v+x \frac{d v}{d x}-\frac{x}{v} e^v\)

⇒ \(\frac{d v}{d x}=\frac{e^v}{v} \Rightarrow e^{-v} v d v=d x\)

Integrating both side 

⇒ e^v (x + c) + 1 + v = 0 

For f(1) = 1 ⇒ c = \(-1-\frac{2}{e} \)

⇒ \(e^v\left(-1-\frac{2}{e}+x\right)+1+v=0 \)

⇒ \(e^{\frac{y}{x}}\left(-1-\frac{2}{e}+x\right)+1+\frac{y}{x}=0 \)

For \( x=a, y=0 \Rightarrow a=\frac{2}{e} \)

⇒ ae = 2 

Calculation

Given

\(\frac{d y}{d x}=\frac{x+y-2}{x-y}\)

Let x = X + h, y = Y + k

⇒ \(\frac{d Y}{d X}=\frac{X+Y}{X-Y}\)

\(\left.\begin{array}{l} \mathrm{h}+\mathrm{k}-2=0 \\ \mathrm{~h}-\mathrm{k}=0 \end{array}\right\} \mathrm{h}=\mathrm{k}=1\)

Let Y = vX

⇒ \(v+\frac{d v}{d X}=\frac{1+v}{1-v} \Rightarrow X-\frac{d v}{d X}=\frac{1+v^2}{1-v}\)

⇒ \(\frac{1-v}{1+v^2} d v=\frac{d X}{X}\)

⇒ \(\tan ^{-1} v-\frac{1}{2} \ln \left(1+v^2\right)=\ln |X|+C\)

⇒ \(\tan ^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{2} \ln \left(1+\left(\frac{y-1}{x-1}\right)^2\right)=\ln |x-1| +C\)

As curve is passing through (2, 1) ⇒ C = 0

⇒ \(\tan ^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{2} \ln \left(1+\left(\frac{y-1}{x-1}\right)^2\right)=\ln |x-1|\)

∴ α = 1 and β = 2

⇒ 5β + α = 11

Calculation

Given

\(\rm y\frac{dx}{dy}=x(\log_e x-\log_e y+1)\)

⇒ \(\rm \frac{dx}{dy}=\frac{x}{y}(\log_e \frac{x}{y}+1)\)

Put \(\frac{x}{y}=v\)

⇒ \(\rm v+y \frac{dv}{dy}=v(\log_e v+1)\)

⇒ \(y \frac{dv}{dy}=v\log_e v\)

⇒ \(\frac{1}{v\log_e v}dv=\frac{dy}{y}\)

⇒ \(\int\frac{dv}{vlog_ev}=\int\frac{dy}{y}\)

⇒ \(log_e|log_ev|=log_e(y) +log_ec\)

⇒ cy = \(|log_e\frac{x}{y}|\)

Put x = e , y = 1 ⇒ c = 1

Given, xy2dy - (x3 + y3)dx = 0 

⇒ xy2dy = (x3 + y3)dx

⇒ \({ dy \over dx } = {x^3 + y^3 \over xy^2}\)

⇒ \({ dy \over dx } = ({x\over y})^2 +{ y \over x}\)

Put y = vx → \({dy \over dx } = v + x {dv \over dx}\)

⇒ \(v + x {dv \over dx} = {1 \over v^2} + v \)

Given, x dy - y dx = \(\sqrt{(x^2+y^2)}dx\)

⇒ x dy = y dx + \(\sqrt{(x^2+y^2)}dx\) 

⇒ \(\frac{dy}{dx}=\frac{y + \sqrt{(x^2+y^2)}}{x}\)

⇒ \(\frac{dy}{dx}= { y \over x}+{ \sqrt{1+({y \over x})^2}}\)

Put y = vx →  \({dy \over dx } = v + x {dv \over dx}\)

⇒ \(v + x {dv \over dx} = v + \sqrt { 1+ v^2}\)

⇒ \(x {dv \over dx} = \sqrt { 1+ v^2}\)

⇒ \( {dv \over\sqrt{ 1+ v^2}} ={dx \over x}\)

Integrating both sides, 

⇒ \(\int {dv \over\sqrt{ 1+ v^2}} =\int{dx \over x}\)d

⇒ \(\log ({v+ \sqrt{ 1+ v^2} }) =\log x + \log c\)

⇒ \( {{y \over x}+ \sqrt{ 1+({y \over x})^2} } =cx\)

⇒ \(y + \sqrt{ x ^2 + y^2} =cx^2\)

∴ The correct answer is option (2).

Given, xy2dy - (x3 + y3)dx = 0 

⇒ xy2dy = (x3 + y3)dx

⇒ \({ dy \over dx } = {x^3 + y^3 \over xy^2}\)

⇒ \({ dy \over dx } = ({x\over y})^2 +{ y \over x}\)

Put y = vx → \({dy \over dx } = v + x {dv \over dx}\)

⇒ \(v + x {dv \over dx} = {1 \over v^2} + v \)

Calculation:

We wish to solve the differential equation

\( \displaystyle \frac{dy}{dx} = -\frac{x^2+3y^2}{3x^2+y^2},\quad y(1)=0. \)

Since both numerator and denominator are homogeneous of degree 2, set

\( y = u\,x,\quad u = \frac{y}{x} \;\longrightarrow\; \frac{dy}{dx} = u + x\frac{du}{dx}. \)

Substituting gives

\( u + x\frac{du}{dx} = -\frac{1+3u^2}{3+u^2} \;\Longrightarrow\; x\frac{du}{dx} = -\frac{1+3u^2}{3+u^2} - u = -\frac{(u+1)^3}{3+u^2}. \)

Separate variables:

\( \displaystyle \int \frac{3+u^2}{(u+1)^3}\,du = -\int \frac{dx}{x}. \)

Notice the partial‐fraction expansion

\( \displaystyle \frac{3+u^2}{(u+1)^3} = \frac{1}{u+1} - \frac{2}{(u+1)^2} + \frac{4}{(u+1)^3}. \)

Hence

\( \displaystyle \int \frac{3+u^2}{(u+1)^3}\,du = \ln|u+1| + \frac{2}{u+1} - \frac{2}{(u+1)^2}. \)

So

\( \displaystyle \ln|u+1| + \frac{2}{u+1} - \frac{2}{(u+1)^2} = -\ln|x| + C. \)

Since \\(u+1 = \frac{x+y}{x} \)), we rewrite in terms of x+y:

\( \displaystyle \ln|x+y| + \frac{2\,x\,y}{(x+y)^2} = C. \)

Apply the initial condition y(1)=0 to get (C=0). Therefore the solution is

\( \displaystyle \ln\lvert x+y\rvert \;+\;\frac{2xy}{(x+y)^2} = 0. \)

Hence, the correct answer is Option 3. 

Calculation

Given equation: \(x\frac{dy}{dx} = y + x \tan(\frac{y}{x})\)

Divide by x: \(\frac{dy}{dx} = \frac{y}{x} + \tan(\frac{y}{x})\)

Let y = vx, then \(\frac{dy}{dx} = v + x\frac{dv}{dx}\)

Substitute in the equation:

\(v + x\frac{dv}{dx} = v + \tan(v)\)

⇒ \(x\frac{dv}{dx} = \tan(v)\)

⇒ \(\frac{dv}{\tan(v)} = \frac{dx}{x}\)

⇒ \(\cot(v) dv = \frac{dx}{x}\)

Integrate both sides:

\(\int \cot(v) dv = \int \frac{dx}{x}\)

⇒ \(\ln|\sin(v)| = \ln|x| + \ln|C|\)

⇒ \(\ln|\sin(v)| = \ln|Cx|\)

Remove the logarithms:

⇒ \(\sin(v) = Cx\)

Substitute v = y/x:

⇒ \(\sin(\frac{y}{x}) = Cx\)

∴ The general solution is \(\sin(\frac{y}{x}) = Cx\).

Hence option 2 is correct

Concept:

If \(\rm\frac{dy}{dx}=f(x,y)\) is a homogeneous differential equation, put y = vx,

Then substitute \(\rm\frac{dy}{dx}=v + x\frac{dv}{dx}\).

Calculation:

Given \(\rm x\cos \frac{y}{x}\left( {y\,dx + x\,dy} \right) = y\sin \frac{y}{x}\left( {x\,dy - y\,dx} \right) \)

⇒ \(\rm xy\cos \frac{y}{x}dx+x^2\cos \frac{y}{x}dy = xy\sin \frac{y}{x}dy-y^2\sin \frac{y}{x}dx\)

⇒ (xy cos\(\rm\frac{y}{x}\) + y² sin \(\rm\frac{y}{x}\))dx = (xy sin\(\rm\frac{y}{x}\) - x² cos \(\rm\frac{y}{x}\))dy

⇒ \(\rm\frac{dy}{dx}=\frac{xy\cos\frac{y}{x}+y^2\sin\frac{y}{x}}{xy\sin\frac{y}{x}-x^2\cos\frac{y}{x}}\)

Dividing the numerator and denominator of R.H.S by x².

⇒ \(\rm\frac{dy}{dx}=\frac{\frac{y}{x}\cos\frac{y}{x}+\frac{y^2}{x^2}\sin\frac{y}{x}}{\frac{y}{x}\sin\frac{y}{x}-\cos\frac{y}{x}}\)

Put y = vx

⇒ \(\rm\frac{dy}{dx}=v+x\frac{dv}{dx}\)

∴ Rewriting the equation, we get:

v + x\(\rm\frac{dv}{dx}\) = \(\rm\frac{v\cos v + v^2\sin v}{v\sin v-\cos v}\)

⇒  x\(\rm\frac{dv}{dx}\) = \(\rm\frac{2v\cos v }{v\sin v-\cos v}\)

⇒ \(\rm\frac{v\sin v-\cos v}{v\cos v}dv=2\frac{dx}{x}\)

Integrating both sides, we get:

⇒ \(\rm\int\frac{v\sin v-\cos v}{v\cos v}dv=2\int\frac{dx}{x}\)

⇒ \(\rm\int(\tan v-\frac{1}{v})dv=2\int\frac{dx}{x}\)

⇒ log |sec v| - log |v| = 2 log |x| + log k.

⇒ log \(\rm\left|\frac{\sec \frac{y}{x}}{\frac{y}{x}}\right|\) = log |x²| + log k.

⇒  log \(\rm\left|\frac{\sec \frac{y}{x}}{\frac{y}{x}}\right|\) - log |x2| = log k.

⇒ log \(\rm\left|\frac{\sec \frac{y}{x}}{\frac{y}{x}\times x^2}\right|\) = log k.

⇒ log \(\rm\left|\frac{\sec \frac{y}{x}}{yx}\right|\) = log k

⇒ \(\rm\frac{\sec \frac{y}{x}}{yx}=k\)

⇒ sec \(\rm\frac{y}{x}\) = kxy

⇒ \(\rm\cos \left( {\frac{y}{x}} \right) = \frac{C}{{xy}} \)