Let \alpha|x|=|y| e^x y-\beta, \alpha, \beta \in N be the solution of the differential equation x ~d y-y ~d x+x y(x ~d y+y ~d x)=0,y(1)=2 . Then \alpha+\beta is equal to \underline\hspace2cm

Differential Equations PYQ — JEE Main 2024 · 8 Apr · Shift 2 · Q57 with Solution

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Correct answer: 4

Given differential equation

We have $x\,dy-y\,dx+xy(x\,dy+y\,dx)=0.$

We also know the solution is of the form $\alpha |x|=|y|e^{xy-\beta}, \qquad \alpha,\beta\in\mathbb N,$ with condition $y(1)=2.$

Rewrite the differential equation

Expand the second term: $xy(x\,dy+y\,dx)=x^2y\,dy+xy^2\,dx.$ So the equation becomes $x\,dy-y\,dx+x^2y\,dy+xy^2\,dx=0.$ Group the $dy$ and $dx$ terms: $x(1+xy)\,dy+y(xy-1)\,dx=0.$

Now divide by $xy$ (valid near the given point $x=1,y=2$ ): $\frac{1+xy}{y}\,dy+\frac{xy-1}{x}\,dx=0.$ That is, $\left(y-\frac1x\right)dx+\left(x+\frac1y\right)dy=0.$

Recognize an exact differential

Notice that $d(xy)=x\,dy+y\,dx,$ and $d(\ln|y|)=\frac{dy}{y}, \qquad d(\ln|x|)=\frac{dx}{x}.$

= d(xy)+d(\ln|y|)-d(\ln|x|).$$ Hence the differential equation becomes $$d\big(xy+\ln|y|-\ln|x|\big)=0.$$ Therefore, $$xy+\ln|y|-\ln|x|=C,$$ for some constant $C$. --- 4. **Convert to the required form** Rearrange: $$\ln\left|\frac{y}{x}\right|=-xy+C.$$ Exponentiating, $$\left|\frac{y}{x}\right|=e^{-xy+C}=e^C e^{-xy}.$$ So $$|y|=|x|e^C e^{-xy}.$$ Thus, $$|y|e^{xy}=e^C|x|.$$ This can be written as $$\alpha |x|=|y|e^{xy-\beta}$$ by taking $$\alpha=e^{C-\beta}.$$ We now use the initial condition to determine the constants in the required natural-number form. --- 5. **Use the condition $y(1)=2$** Substitute $x=1$, $y=2$ into $$xy+\ln|y|-\ln|x|=C.$$ Then $$C=(1)(2)+\ln 2-\ln 1=2+\ln 2.$$ So $$e^C=e^{2+\ln 2}=2e^2.$$ Hence the solution is $$|y|e^{xy}=2e^2|x|.$$ Equivalently, $$2|x|=|y|e^{xy-2}.$$ Comparing with $$\alpha |x|=|y|e^{xy-\beta},$$ we get $$\alpha=2,\qquad \beta=2.$$ Therefore, $$\alpha+\beta=2+2=4.$$ --- 6. **Comparison with stored answer** Stored correct answer: $4$ Our derived answer is also $4$, so it agrees.

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View written solutionFree

Correct answer: 4

Given differential equation

We have $x\,dy-y\,dx+xy(x\,dy+y\,dx)=0.$

We also know the solution is of the form $\alpha |x|=|y|e^{xy-\beta}, \qquad \alpha,\beta\in\mathbb N,$ with condition $y(1)=2.$

Rewrite the differential equation

Expand the second term: $xy(x\,dy+y\,dx)=x^2y\,dy+xy^2\,dx.$ So the equation becomes $x\,dy-y\,dx+x^2y\,dy+xy^2\,dx=0.$ Group the $dy$ and $dx$ terms: $x(1+xy)\,dy+y(xy-1)\,dx=0.$

Now divide by $xy$ (valid near the given point $x=1,y=2$ ): $\frac{1+xy}{y}\,dy+\frac{xy-1}{x}\,dx=0.$ That is, $\left(y-\frac1x\right)dx+\left(x+\frac1y\right)dy=0.$

Recognize an exact differential

Notice that $d(xy)=x\,dy+y\,dx,$ and $d(\ln|y|)=\frac{dy}{y}, \qquad d(\ln|x|)=\frac{dx}{x}.$

= d(xy)+d(\ln|y|)-d(\ln|x|).$$ Hence the differential equation becomes $$d\big(xy+\ln|y|-\ln|x|\big)=0.$$ Therefore, $$xy+\ln|y|-\ln|x|=C,$$ for some constant $C$. --- 4. **Convert to the required form** Rearrange: $$\ln\left|\frac{y}{x}\right|=-xy+C.$$ Exponentiating, $$\left|\frac{y}{x}\right|=e^{-xy+C}=e^C e^{-xy}.$$ So $$|y|=|x|e^C e^{-xy}.$$ Thus, $$|y|e^{xy}=e^C|x|.$$ This can be written as $$\alpha |x|=|y|e^{xy-\beta}$$ by taking $$\alpha=e^{C-\beta}.$$ We now use the initial condition to determine the constants in the required natural-number form. --- 5. **Use the condition $y(1)=2$** Substitute $x=1$, $y=2$ into $$xy+\ln|y|-\ln|x|=C.$$ Then $$C=(1)(2)+\ln 2-\ln 1=2+\ln 2.$$ So $$e^C=e^{2+\ln 2}=2e^2.$$ Hence the solution is $$|y|e^{xy}=2e^2|x|.$$ Equivalently, $$2|x|=|y|e^{xy-2}.$$ Comparing with $$\alpha |x|=|y|e^{xy-\beta},$$ we get $$\alpha=2,\qquad \beta=2.$$ Therefore, $$\alpha+\beta=2+2=4.$$ --- 6. **Comparison with stored answer** Stored correct answer: $4$ Our derived answer is also $4$, so it agrees.

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