a) Evaluate: x^2x^4 2x^2 15 dx . [4] b) Solve: Sin x dydx + y x = x x . [4] OR Solve: (1+x^2) dydx + y = e^ ^ 1x

a) Evaluate: $\int \frac{x^2}{x^4-2x^2-15}dx$

Step 1: Factorize the denominator

The denominator is $x^4-2x^2-15$. Let $u=x^2$, then the expression becomes: $$u^2-2u-15=(u-5)(u+3)$$ Substituting back $u=x^2$, we get: $$x^4-2x^2-15=(x^2-5)(x^2+3)$$

Step 2: Partial fraction decomposition

We express the integrand as partial fractions: $$\frac{x^2}{(x^2-5)(x^2+3)}=\frac{Ax+B}{x^2-5}+\frac{Cx+D}{x^2+3}$$ Multiply both sides by $(x^2-5)(x^2+3)$: $$x^2=(Ax+B)(x^2+3)+(Cx+D)(x^2-5)$$ Expand and collect like terms: $$x^2=A{x}^3+3Ax+B{x}^2+3B+C{x}^3-5Cx+D{x}^2-5D$$ $$x^2=(A+C)x^3+(B+D)x^2+(3A-5C)x+(3B-5D)$$ Equate coefficients: $$\{\begin{array}{l}A+C=0\\ B+D=1\\ 3A-5C=0\\ 3B-5D=0\end{array}$$ Solve the system: From $A+C=0$, we get $C=-A$. Substitute $C=-A$ into $3A-5C=0$: $$3A-5(-A)=0⟹3A+5A=0⟹8A=0⟹A=0$$ Thus, $C=0$. From $B+D=1$ and $3B-5D=0$, solve: $$3B-5(1-B)=0⟹3B-5+5B=0⟹8B=5⟹B=\frac{5}{8}$$ $$D=1-B=1-\frac{5}{8}=\frac{3}{8}$$ So, the partial fractions are: $$\frac{x^2}{(x^2-5)(x^2+3)}=\frac{5/8}{x^2-5}+\frac{3/8}{x^2+3}$$

Step 3: Rewrite the integral

$$\int \frac{x^2}{x^4-2x^2-15}dx=\frac{5}{8}\int \frac{1}{x^2-5}dx+\frac{3}{8}\int \frac{1}{x^2+3}dx$$

Step 4: Evaluate each integral separately

1. For $\int \frac{1}{x^2-5}dx$: $$\int \frac{1}{x^2-5}dx=\frac{1}{2\sqrt{5}}\ln \left|\frac{x-\sqrt{5}}{x+\sqrt{5}}\right|+C_1$$

2. For $\int \frac{1}{x^2+3}dx$: $$\int \frac{1}{x^2+3}dx=\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x}{\sqrt{3}}\right)+C_2$$

Step 5: Combine the results

$$\int \frac{x^2}{x^4-2x^2-15}dx=\frac{5}{8}\cdot \frac{1}{2\sqrt{5}}\ln \left|\frac{x-\sqrt{5}}{x+\sqrt{5}}\right|+\frac{3}{8}\cdot \frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x}{\sqrt{3}}\right)+C$$ Simplify: $$=\frac{5}{16\sqrt{5}}\ln \left|\frac{x-\sqrt{5}}{x+\sqrt{5}}\right|+\frac{\sqrt{3}}{8}{\tan }^{-1}\left(\frac{x}{\sqrt{3}}\right)+C$$ Rationalize $\frac{5}{16\sqrt{5}}$: $$\frac{5}{16\sqrt{5}}=\frac{5\sqrt{5}}{16\cdot 5}=\frac{\sqrt{5}}{16}$$ Thus, the final answer is: $$\boxed{\frac{\sqrt{5}}{16}\ln \left|\frac{x-\sqrt{5}}{x+\sqrt{5}}\right|+\frac{\sqrt{3}}{8}{\tan }^{-1}\left(\frac{x}{\sqrt{3}}\right)+C}$$

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b) Solve: $\sin x\frac{dy}{dx}+y\cos x=x\sin x$

Step 1: Rewrite the differential equation

$$\sin x\frac{dy}{dx}+y\cos x=x\sin x$$ Divide both sides by $\sin x$ (assuming $\sin x\ne 0$): $$\frac{dy}{dx}+y\cot x=x$$

Step 2: Identify the integrating factor

The equation is linear in $y$ of the form: $$\frac{dy}{dx}+P(x)y=Q(x)$$ where $P(x)=\cot x$ and $Q(x)=x$.

The integrating factor (IF) is: $$\text{IF}=e^{\int P(x)dx}=e^{\int \cot xdx}$$ Compute $\int \cot xdx$: $$\int \cot xdx=\int \frac{\cos x}{\sin x}dx=\ln |\sin x|$$ Thus: $$\text{IF}=e^{\ln |\sin x|}=\sin x$$

Step 3: Multiply through by the integrating factor

$$\sin x\frac{dy}{dx}+y\cos x=x\sin x$$ The left side is the derivative of $y\sin x$: $$\frac{d}{dx}(y\sin x)=x\sin x$$

Step 4: Integrate both sides

$$\int \frac{d}{dx}(y\sin x)dx=\int x\sin xdx$$ $$y\sin x=\int x\sin xdx$$

Step 5: Evaluate $\int x\sin xdx$ using integration by parts

Let $u=x$, $dv=\sin xdx$, then $du=dx$, $v=-\cos x$: $$\int x\sin xdx=-x\cos x+\int \cos xdx=-x\cos x+\sin x+C$$

Step 6: Solve for $y$

$$y\sin x=-x\cos x+\sin x+C$$ $$y=\frac{-x\cos x+\sin x+C}{\sin x}$$ $$y=-x\cot x+1+C\csc x$$

Final Answer:

$$\boxed{y=-x\cot x+1+C\csc x}$$

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OR

Solve: $(1+x^2)\frac{dy}{dx}+y=e^{{\tan }^{-1}x}$

Step 1: Rewrite the differential equation

$$(1+x^2)\frac{dy}{dx}+y=e^{{\tan }^{-1}x}$$ Divide by $(1+x^2)$: $$\frac{dy}{dx}+\frac{y}{1+x^2}=\frac{e^{{\tan }^{-1}x}}{1+x^2}$$

Step 2: Identify the integrating factor

The equation is linear in $y$ of the form: $$\frac{dy}{dx}+P(x)y=Q(x)$$ where $P(x)=\frac{1}{1+x^2}$ and $Q(x)=\frac{e^{{\tan }^{-1}x}}{1+x^2}$.

The integrating factor (IF) is: $$\text{IF}=e^{\int P(x)dx}=e^{\int \frac{1}{1+x^2}dx}=e^{{\tan }^{-1}x}$$

Step 3: Multiply through by the integrating factor

$$e^{{\tan }^{-1}x}\frac{dy}{dx}+\frac{e^{{\tan }^{-1}x}}{1+x^2}y=\frac{e^{{\tan }^{-1}x}\cdot e^{{\tan }^{-1}x}}{1+x^2}$$ The left side is the derivative of $y{e}^{{\tan }^{-1}x}$: $$\frac{d}{dx}\left(y{e}^{{\tan }^{-1}x}\right)=\frac{e^{2{\tan }^{-1}x}}{1+x^2}$$

Step 4: Integrate both sides

$$\int \frac{d}{dx}\left(y{e}^{{\tan }^{-1}x}\right)dx=\int \frac{e^{2{\tan }^{-1}x}}{1+x^2}dx$$ Let $u={\tan }^{-1}x$, then $du=\frac{1}{1+x^2}dx$: $$y{e}^{{\tan }^{-1}x}=\int e^{2u}du=\frac{1}{2}{e}^{2u}+C=\frac{1}{2}{e}^{2{\tan }^{-1}x}+C$$

Step 5: Solve for $y$

$$y=\frac{1}{2}{e}^{{\tan }^{-1}x}+C{e}^{-{\tan }^{-1}x}$$

Final Answer:

$$\boxed{y=\frac{1}{2}{e}^{{\tan }^{-1}x}+C{e}^{-{\tan }^{-1}x}}$$