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Exercise 9.3 · Q12

Q.Solve the following differential equation: x(x2−1)dydx=1;y=0x(x^2 - 1) \frac{dy}{dx} = 1; y = 0 when x=2x = 2

Puducherry CbseNCERTSubjective· 3mImportance★★★★★
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Separate, integrate by partial fractions, and use y(2)=0y(2)=0:   y=12log⁡ ⁣(4(x2−1)3x2).\;y=\frac{1}{2}\log\!\left(\frac{4(x^2-1)}{3x^2}\right).

1. Separate the variables

x(x2−1)dydx=1  ⇒  dy=dxx(x2−1)=dxx(x−1)(x+1).x(x^2-1)\frac{dy}{dx}=1\;\Rightarrow\;dy=\frac{dx}{x(x^2-1)}=\frac{dx}{x(x-1)(x+1)}.

2. Partial fractions

Write 1x(x−1)(x+1)=Ax+Bx−1+Cx+1\dfrac{1}{x(x-1)(x+1)}=\dfrac{A}{x}+\dfrac{B}{x-1}+\dfrac{C}{x+1}. Covering each factor and substituting its root:

A=1(−1)(1)=−1,B=1(1)(2)=12,C=1(−1)(−2)=12.A=\frac{1}{(-1)(1)}=-1,\quad B=\frac{1}{(1)(2)}=\frac12,\quad C=\frac{1}{(-1)(-2)}=\frac12.

3. Integrate

y=∫(−1x+12(x−1)+12(x+1))dx=−log⁡∣x∣+12log⁡∣x−1∣+12log⁡∣x+1∣+C.y=\int\left(-\frac{1}{x}+\frac{1}{2(x-1)}+\frac{1}{2(x+1)}\right)dx=-\log|x|+\tfrac12\log|x-1|+\tfrac12\log|x+1|+C.

Combine the last two logs, 12log⁡∣x−1∣+12log⁡∣x+1∣=12log⁡∣x2−1∣\tfrac12\log|x-1|+\tfrac12\log|x+1|=\tfrac12\log|x^2-1|, so

y=12log⁡∣x2−1x2∣+C.y=\tfrac12\log\left|\frac{x^2-1}{x^2}\right|+C.

4. Use the condition y=0y=0 at x=2x=2 …

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