Differential Equations Question 47
Question: $ y+x^{2}=\frac{dy}{dx} $ has the solution
[EAMCET 2002]
Options:
A) $ y+x^{2}+2x+2=ce^{x} $
B) $ y+x+x^{2}+2=ce^{2x} $
C) $ y+x+2x^{2}+2=ce^{x} $
D) $ y^{2}+x+x^{2}+2=ce^{x} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ y+x^{2}=\frac{dy}{dx} $
Therefore $ \frac{dy}{dx}-y=x^{2} $
This is the linear differential equation in y, where $ P=-1,Q=x^{2} $
I.F. $ ={e^{\int{P.dx}}} $
$ ={e^{\int{-dx}}}={e^{-x}} $
Hence solution, $ y.(I\text{.F}).=\int{Q.(I\text{.F})dx+c} $
Therefore $ y{e^{-x}}=-x^{2}{e^{-x}}-2x{e^{-x}}-2{e^{-x}}+c $
Therefore $ y+x^{2}+2x+2=ce^{x} $ .
BETA
