(a) Resolve into partial fractions $\dfrac{2x}{(x^2+1)(x-1)}$. **OR** (b) If $y=e^{\tan^{-1}x}$, show that $(1+x^2)y''+(2x-1)y'=0$.

(a) Resolve into partial fractions (2x)/((x²+1)(x-1)). **OR** (b) If y=e^tan⁻¹x, show that (1+x²)y''+(2x-1)y'=

Question 117 of 128

\dfrac{2x}{(x^2+1)(x-1)}

(b) If

y=e^{\tan^{-1}x}

, show that

(1+x^2)y''+(2x-1)y'=0

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2024Subjective· 5mImportance★★★★★

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$\dfrac{2x}{(x^2+1)(x-1)}=\dfrac{1-x}{x^2+1}+\dfrac{1}{x-1}$ .

Let

$\dfrac{2x}{(x^2+1)(x-1)}=\dfrac{Ax+B}{x^2+1}+\dfrac{C}{x-1}.$

Multiplying both sides by $(x^2+1)(x-1)$ :

$2x=(Ax+B)(x-1)+C(x^2+1).$

Put $x=1$ : $2=C(1^2+1)=2C\ \Rightarrow\ C=1$ .

Expand the right side: $(Ax+B)(x-1)+C(x^2+1)=Ax^2-Ax+Bx-B+Cx^2+C=(A+C)x^2+(B-A)x+(C-B)$ .

Matching with $2x=0\cdot x^2+2x+0$ :

$x^2$ coefficient: $A+C=0\ \Rightarrow\ A=-C=-1$ .

$x$ coefficient: $B-A=2\ \Rightarrow\ B=2+A=2-1=1$ .

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