If the solution curve, of the differential equation \fracd y~d x=\fracx+y-2x-y passing through the point (2,1) is \tan ^-1(\fracy-1x-1)-\frac1\beta \log _e(\alpha+(\fracy-1x-1)^2)=\log _e|x-1| , then 5 \beta+\alpha is equal to \underline\hspace2cm .

Differential Equations PYQ — JEE Main 2024 · 27 Jan · Shift 2 · Q60 with Solution

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Correct answer: 11

Shift the origin to simplify the differential equation

Given $\frac{dy}{dx}=\frac{x+y-2}{x-y}$ and the curve passes through $(2,1)$ .

Notice the constants suggest the substitution $X=x-1, \qquad Y=y-1.$ Then $y=Y+1, \quad x=X+1, \quad \frac{dy}{dx}=\frac{dY}{dX}.$

Now transform the DE: $x+y-2=(X+1)+(Y+1)-2=X+Y,$ $x-y=(X+1)-(Y+1)=X-Y.$ So the equation becomes $\frac{dY}{dX}=\frac{X+Y}{X-Y}.$

Use the homogeneous substitution

Let $Y=vX \quad \Rightarrow \quad \frac{dY}{dX}=v+X\frac{dv}{dX}.$ Substitute into the DE: $v+X\frac{dv}{dX}=\frac{X+vX}{X-vX}=\frac{1+v}{1-v}.$ Hence $X\frac{dv}{dX}=\frac{1+v}{1-v}-v.$ Simplify: $\frac{1+v}{1-v}-v=\frac{1+v-v(1-v)}{1-v}= rac{1+v-v+v^2}{1-v}= rac{1+v^2}{1-v}.$ Therefore $X\frac{dv}{dX}=\frac{1+v^2}{1-v}.$ So $\frac{1-v}{1+v^2}\,dv=\frac{dX}{X}.$

Integrate both sides

Integrate: $\int \frac{1-v}{1+v^2}\,dv=\int \frac{dX}{X}.$ Now

=\int \frac{1}{1+v^2}\,dv-\int \frac{v}{1+v^2}\,dv.$$ So $$=\tan^{-1}v-\frac{1}{2}\ln(1+v^2).$$ Thus, $$\tan^{-1}v-\frac{1}{2}\ln(1+v^2)=\ln|X|+C.$$ Substitute back $$v=\frac{Y}{X}=\frac{y-1}{x-1}, \qquad X=x-1.$$ Hence $$\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.$$ --- 4. **Use the point $(2,1)$ to determine the constant** At $(x,y)=(2,1)$, $$\frac{y-1}{x-1}=\frac{0}{1}=0.$$ Then left side becomes $$\tan^{-1}(0)-\frac{1}{2}\ln(1+0)=0-0=0,$$ and right side becomes $$\ln|2-1|+C=\ln 1 + C=0+C=C.$$ So $$C=0.$$ Therefore the required solution curve is $$\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|.$$ --- 5. **Compare with the given form** Given form: $$\tan^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{\beta}\ln\left(\alpha+\left(\frac{y-1}{x-1}\right)^2\right)=\ln|x-1|.$$ Comparing, $$\alpha=1, \qquad \frac{1}{\beta}=\frac{1}{2} \Rightarrow \beta=2.$$ Hence $$5\beta+\alpha=5(2)+1=11.$$ --- 6. **Final answer** $$\boxed{11}$$

More from Differential Equations

View written solutionFree

Correct answer: 11

Shift the origin to simplify the differential equation

Given $\frac{dy}{dx}=\frac{x+y-2}{x-y}$ and the curve passes through $(2,1)$ .

Notice the constants suggest the substitution $X=x-1, \qquad Y=y-1.$ Then $y=Y+1, \quad x=X+1, \quad \frac{dy}{dx}=\frac{dY}{dX}.$

Now transform the DE: $x+y-2=(X+1)+(Y+1)-2=X+Y,$ $x-y=(X+1)-(Y+1)=X-Y.$ So the equation becomes $\frac{dY}{dX}=\frac{X+Y}{X-Y}.$

Use the homogeneous substitution

Integrate both sides

Integrate: $\int \frac{1-v}{1+v^2}\,dv=\int \frac{dX}{X}.$ Now

=\int \frac{1}{1+v^2}\,dv-\int \frac{v}{1+v^2}\,dv.$$ So $$=\tan^{-1}v-\frac{1}{2}\ln(1+v^2).$$ Thus, $$\tan^{-1}v-\frac{1}{2}\ln(1+v^2)=\ln|X|+C.$$ Substitute back $$v=\frac{Y}{X}=\frac{y-1}{x-1}, \qquad X=x-1.$$ Hence $$\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.$$ --- 4. **Use the point $(2,1)$ to determine the constant** At $(x,y)=(2,1)$, $$\frac{y-1}{x-1}=\frac{0}{1}=0.$$ Then left side becomes $$\tan^{-1}(0)-\frac{1}{2}\ln(1+0)=0-0=0,$$ and right side becomes $$\ln|2-1|+C=\ln 1 + C=0+C=C.$$ So $$C=0.$$ Therefore the required solution curve is $$\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|.$$ --- 5. **Compare with the given form** Given form: $$\tan^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{\beta}\ln\left(\alpha+\left(\frac{y-1}{x-1}\right)^2\right)=\ln|x-1|.$$ Comparing, $$\alpha=1, \qquad \frac{1}{\beta}=\frac{1}{2} \Rightarrow \beta=2.$$ Hence $$5\beta+\alpha=5(2)+1=11.$$ --- 6. **Final answer** $$\boxed{11}$$

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