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Phrases Differential Equation PYQ


Phrases PYQ
The curve satisfying the differential equation ydx-(x+3y2)dy=0 and passing through the point (1,1) also passes through the point __________





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Phrases Previous Year PYQPhrases NIMCET 2019 PYQ

Solution

Given: \(y\,dx-(x+3y^{2})\,dy=0 \;\Rightarrow\; y\,\dfrac{dx}{dy}=x+3y^{2}\).

So \( \dfrac{dx}{dy}-\dfrac{1}{y}x=3y\) (linear). Integrating factor \(=\exp\!\int\!-\dfrac{1}{y}dy=\dfrac{1}{y}\).

\(\displaystyle \frac{d}{dy}\!\left(\frac{x}{y}\right)=3 \;\Rightarrow\; \frac{x}{y}=3y+C \;\Rightarrow\; x=3y^{2}+Cy.\)

Through \((1,1)\): \(1=3(1)+C(1)\Rightarrow C=-2\). Hence curve: \(x=3y^{2}-2y\).


Phrases PYQ
The solution of (ex + 1) y dy = (y + 1) edx is





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Phrases Previous Year PYQPhrases NIMCET 2017 PYQ

Solution

Question: Solve $(e^x+1)\,y\,dy=(y+1)\,e^x\,dx$.

Solution:

Separate: $\dfrac{y}{y+1}\,dy=\dfrac{e^x}{e^x+1}\,dx$

Integrate: $y-\ln(y+1)=\ln(e^x+1)+C$

Answer (implicit): $\boxed{\,y-\ln(1+y)=\ln(1+e^x)+C\,}$

Equivalently: $\dfrac{e^{y}}{y+1}=K(1+e^x)$.

Answer : $e^y=k(y+1)(1+e^x)$


Phrases PYQ
The solution of the differential equation $\dfrac{dy}{dx}=e^{x+y}+x^2e^y$ is





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Phrases Previous Year PYQPhrases NIMCET 2017 PYQ

Solution

$\dfrac{dy}{dx}=e^y(e^x+x^2)\ \Rightarrow\ e^{-y}dy=(e^x+x^2)dx$

$\int e^{-y}dy=\int (e^x+x^2)dx$ $\Rightarrow\ -e^{-y}=e^x+\dfrac{x^3}{3}+C$

Hence, $e^{-y}+e^x+\dfrac{x^3}{3}=C$ (or $y=-\ln\!\big(C-e^x-\dfrac{x^3}{3}\big)$).

Answer: $\boxed{e^{-y}+e^x+\dfrac{x^3}{3}=C}$ ✅



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