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NIMCET Previous Year Questions (PYQs)
NIMCET Differential Equation PYQ
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NIMCET Previous Year PYQNIMCET 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: \(x=3y^{2}-2y\).
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NIMCET Previous Year PYQNIMCET 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 : $e^y=k(y+1)(1+e^x)$
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NIMCET Previous Year PYQNIMCET 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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NIMCET
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