Question

Consider the differential equation: dy/dx = (y-2)(x^2+1)

Find the value of d^2y/dx^2 at the point (1,3)

Answer 1

Answer:

6

Step-by-step explanation:

Given,

$\dfrac{dy}{dx}=(y-2)(x^2+1)$ ...(i)

Differentiating w.r. to x.

$\dfrac{d^2y}{dx^2}= (y-2)(2x)+ (x^2+1)\dfrac{dy}{dx}$

From equation (1)

$\dfrac{d^2y}{dx^2}= (y-2)(2x)+ (x^2+1)^2(y-2)$

Now, at the point (1,3)

$\dfrac{d^2y}{dx^2}= (3-2)(2\times 1)+ (1^2+1)^2(3-2)$

$\dfrac{d^2y}{dx^2}=6$

Answer 2

The required value is  $\dfrac{d^2y}{dx^2}_{(1,3)}=6$.

Important information:

• The given differential equation is $\dfrac{dy}{dx}=(y-2)(x^2+1)$.

Differential equation:

Differentiate the given differential equation with respect to $x$.

$\dfrac{d^2y}{dx^2}=(y-2)\dfrac{d}{dx}(x^2+1)+(x^2+1)\dfrac{d}{dx}(y-2)$

$\dfrac{d^2y}{dx^2}=(y-2)(2x)+(x^2+1)\dfrac{dy}{dx}$

$\dfrac{d^2y}{dx^2}=2x(y-2)+(x^2+1)[(y-2)(x^2+1)]$

$\dfrac{d^2y}{dx^2}=2x(y-2)+(x^2+1)^2(y-2)$

Subsitute $x=1,y=3$.

$\dfrac{d^2y}{dx^2}_{(1,3)}=2(1)(3-2)+((1)^2+1)^2(3-2)$

$\dfrac{d^2y}{dx^2}_{(1,3)}=2(1)+(1+1)^2(1)$

$\dfrac{d^2y}{dx^2}_{(1,3)}=2+4$

$\dfrac{d^2y}{dx^2}_{(1,3)}=6$

Therefore, the required value is $\dfrac{d^2y}{dx^2}_{(1,3)}=6$.

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