Fun With Calculus-6

Calculus Level 5

If the first four terms of the power series expansion of the solution of the differential equation y ′ ′ − ( 1 + x 2 ) y = 0 \large y''-(1+x^2) y=0 satisfying the initial conditions y ∣ x = 0 = − 2 , y ′ ∣ x = 0 = 2 y|_{x=0}=-2,y'|_{x=0}=2 is

a + b x + c x 2 + d x 3 . a+b x+c x^2+d x^3. . Then the value of a + b + c + d a+b+c+d is

Try my fun series


The answer is -0.66.

Akshay Sharma by Akshay Sharma , 21, India

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2 solutions

y ′ ′ ( x ) = ( 1 + x 2 ) y ( x ) y''(x) = (1+x^{2})y(x)
Putting x = 0 x = 0
y ′ ′ ( 0 ) = ( 1 + 0 ) y ( 0 ) = − 2 y''(0) = (1+0)y(0) = -2
Differentiating the equation,
y ′ ′ ′ ( x ) = ( 1 + x 2 ) y ′ ( x ) + 2 x y ( x ) y'''(x) = (1+x^{2})y'(x) + 2xy(x)
Put x = 0,
y ′ ′ ′ ( x ) = y ′ ( x ) = 2 y'''(x) = y'(x) = 2
The first four terms of the power series are,
y ( 0 ) + y ′ ( 0 ) x + y ′ ′ ( 0 ) 2 ! x 2 + y ′ ′ ′ ( 0 ) 3 ! x 3 = − 2 + 2 x − x 2 + 1 3 x 3 y(0) + y'(0)x + \dfrac{y''(0)}{2!}x^{2} + \dfrac{y'''(0)}{3!}x^{3} = -2 + 2x - x^{2} + \dfrac{1}{3}x^{3}
a + b + c + d = − 2 + 2 − 1 + 1 3 = − 2 3 ≈ − 0.66 a + b + c + d = -2+2-1+\dfrac{1}{3} = -\dfrac{2}{3} \approx -0.66


8 Helpful 1 Interesting 0 Brilliant 0 Confused
Samarth Vashisht
Mar 2, 2019

Let F(x)=a+bx+cx^2+dx^3+.....

Given: F(0)=-2; =>a=-2 

       F'(x)=2; =>b=2

Now, F''(x)=2c+6dx+.....=(1+x^2)(a+bx+cx^2+.....) compare coefficients, 2c=a;=>c=-1 6d=b;=>d=1/3

Therefore, a+b+c+d=-2+2-1+1/3 =-0.667

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