The solution of $y' = 1 + x + {y^2} + x{y^2},y\left( 0 \right) = 0$ is

${y^2} = \exp \left( {x + \frac{{{x^2}}}{2}} \right) - 1$

${y^2} = 1 + c\exp \left( {x + \frac{{{x^2}}}{2}} \right)$

$y = \tan \left( {c + x + {x^2}} \right)$

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

The solution of

y' = 1 + x + {y^2} + x{y^2},y\left( 0 \right) = 0

View answer and explanation

Correct option: D

y = \tan \left( {x + \frac{{{x^2}}}{2}} \right)

Option D,

y = \tan \left( {x + \frac{{{x^2}}}{2}} \right)

, is the correct solution because it satisfies the given differential equation and the initial condition

y(0) = 0

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)