JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

A function

y=f(x)

satisfies

(x+1) \cdot f^{\prime}(x)-2\left(x^2+x\right) f(x)=\frac{e^{x^2}}{(x+1)}, \forall x>-1

.
If

f(0)=5

, then

f(x)

View answer and explanation

Correct option: B

\left(\frac{6 x+5}{x+1}\right) \cdot e^{x^2}

Option B (

\left(\frac{{6 x+5}}{{x+1}}\right) \cdot e^{x^2}

) is correct because it satisfies the given differential equation and the initial condition f(0)=5.

More Questions from DIFFERENTIAL EQUATIONS

Practice All Chapter MCQs →

y=y(x)

\left(3 y^{2}-5 x^{2}\right) y d x+2 x\left(x^{2}-y^{2}\right) d y=0

y(1)=1

\left|(y(2))^{3}-12 y(2)\right|

x

{x^2} + {y^2} = 1

\left( {y' = \frac{{dy}}{{dx}},y'' = \frac{{{d^2}y}}{{d{x^2}}}} \right)

x = A\cos 4t + B\sin 4t,

\frac{{{d^2}x}}{{d{t^2}}}

y=y_{1}(x)

y=y_{2}(x)

\frac{d y}{d x}=y+7

y_{1}(0)=0

y_{2}(0)=1

y=y_{1}(x)

y=y_{2}(x)

y=y(t)

\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}

\alpha>0, \beta>0

\gamma>0

\lim \limits_{{x \rightarrow \infty}} y(t)