3 Marks

Question 13 Marks

Solve the following differential equation :
$
\left(x^2-1\right) \frac{d y}{d x}+2 x y=\frac{1}{x^2-1},|x| \neq 1
$

Answer

From the given differential equation
$
\begin{aligned}
\left(x^2-1\right) \frac{d y}{d x}+2 x y & =\frac{1}{x^2-1} \\
\Rightarrow \quad \frac{d y}{d x}+\frac{2 x}{x^2-1} y & =\frac{1}{\left(x^2-1\right)^2}
\end{aligned}
$
Comparing equation (1) with $\frac{d y}{d x}+ P y= Q$,
$
P=\frac{2 x}{x^2-1} \text { and } Q=\frac{1}{\left(x^2-1\right)^2}
$
Integrating factor I.F. $=e^{\int \frac{2 x}{x^2-1} d x}$
Let$
x^2-1=t
$
$
\begin{array}{ll}
\therefore & 2 x d x=d t \\
\therefore & \text { I.F. }=e^{\int \frac{d t}{t}}=e^{\int \log t}=t=\left(x^2-1\right)
\end{array}
$
Hence the solution of this differential equation will be :
$
\begin{array}{rlrl}
& y . I . F . =\int(I . F .)(Q) d x \\
\Rightarrow & y\left(x^2-1\right) =\int\left(x^2-1\right) \cdot \frac{1}{\left(x^2-1\right)^2} d x+C \\
\Rightarrow & y\left(x^2-1\right) =\int \frac{d x}{x^2-1}+C \\
\Rightarrow & y\left(x^2-1\right) =\frac{1}{2} \log \left|\frac{x-1}{x+1}\right|+C \\
\therefore & y =\frac{1}{2\left(x^2-1\right)} \log \left|\frac{x-1}{x+1}\right|+\frac{C}{x^2-1}
\end{array}
$

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Question 23 Marks

Solve : $\cos (x+y) d y=d x$.

Answer

Writing the given equation in the following form:
$
\frac{d y}{d x}=\frac{1}{\cos (x+y)}
$
Here it is clear that in equation 1 , we cannot separate the variables $x$ and $y$ until we substitute $x+y=t$.
Hence taking $x+y=t$
$
\therefore \quad 1+\frac{d y}{d x}=\frac{d t}{d x} \therefore \frac{d y}{d x}=\frac{d t}{d x}-1
$
Hence from equation (1),
$
\begin{array}{l}
\frac{d t}{d x}-1=\frac{1}{\cos t} \\
\Rightarrow \quad \frac{d t}{d x}=\frac{1}{\cos t}+1=\frac{1+\cos t}{\cos t}
\end{array}
$
Now we shall separate the variables
$
\frac{\cos t}{1+\cos t} d t=d x
$
Integrating both sides of equation (2),
$
\begin{array}{l}
\int d x=\int \frac{\cos t}{1+\cos t} d t \\
\int d x=\int\left(1-\frac{1}{1+\cos t}\right) d t
\end{array}
$
$
\begin{array}{rlrl}
\Rightarrow & & \int d x & =\int d t-\int \frac{1}{2 \cos ^2 t / 2} d t \\
\Rightarrow & & \int d x & =\int d t-\frac{1}{2} \int \sec ^2 t / 2 d t \\
\Rightarrow & x & =t-\tan t / 2+C_1
\end{array}
$
Putting the value of $t$
$
\begin{aligned}
x & =x+y-\tan \left(\frac{x+y}{2}\right)+C_1 \\
\Rightarrow \quad y & =\tan \left(\frac{x+y}{2}\right)-C_1
\end{aligned}
$
Hence$
y=\tan \left(\frac{x+y}{2}\right)+C \text { where } C=-C_1
$

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Maths 3 Marks

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