Question

If the value of $\left( {{x^2} + 1} \right)\dfrac{{{d^2}y}}{{d{x^2}}} = 2x\dfrac{{dy}}{{dx}}$, where y’(0) =3 & y(0)=1, then the value of y(1) is equal to:

Answer 1

Hint: Start by finding the value of $\dfrac{{y''}}{{y'}}$ from the given equation and then integrate the equation using a suitable formula and find out the required value that is y(1).

Complete Step-by-Step solution:

Let us denote $\dfrac{{{d^2}y}}{{d{x^2}}}$ by y’’ and $\dfrac{{dy}}{{dx}}$ by y’, therefore first we are going to find the value of $\dfrac{{y''}}{{y'}}$, therefore,

$\dfrac{{y''}}{{y'}} = \dfrac{{2x}}{{{x^2} + 1}}$

Now the next step is to integrate, on integrating we get,

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

Above we have used the formula,

$\int {\dfrac{{2a}}{{{a^2} + 1}}} da = ?$

${a^2} + 1 = u$,

On differentiating both sides, we get,

$2a\,da = du$

Therefore,

$\int {\dfrac{{2a}}{{{a^2} + 1}}} da = \int {\dfrac{{du}}{u}} = \ln \,u + \ln \,C$

And we know that, ln a+ln b=ln (ab), therefore,

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

It is given to us that,

  y’(0)=3

Therefore, if we equate the value to the equation, we get,

  c=3,

Putting c=3 in the equation, we get,

$\,y' = \,\left( {{x^2} + 1} \right)3$

The value of y from the above becomes,

$\,y = \,{x^3} + 3x + 1$

On putting the value of x=1 we obtain the value of y(1), therefore,

The value of y(1) is 5.

Note: In this question for integration we have used the substitution method which is one of the easier methods to perform integration and also takes less time. Always remembers the basic formulas of integration.