a) Which type of differential equation x dydx+y x=x x represents? Also solve it. [3] b) Evaluate: x(x 1)(x^2+1)\, dx . [3] c) Two cars start from certain places at the same instant. One goes east at…

a) Type and Solution of the Differential Equation

The given differential equation is: $$\sin x\frac{dy}{dx}+y\cos x=x\sin x$$

Type: This is a First Order Linear Differential Equation in the standard form $\frac{dy}{dx}+P(x)y=Q(x)$.

Solution:

1. Divide the entire equation by $\sin x$ to get the standard form: $$\frac{dy}{dx}+y\frac{\cos x}{\sin x}=x$$ $$\frac{dy}{dx}+y\cot x=x$$ Here, $P(x)=\cot x$ and $Q(x)=x$.

2. Find the Integrating Factor (I.F.): $$\text{I.F.}=e^{\int P(x)dx}=e^{\int \cot xdx}=e^{\ln (\sin x)}=\sin x$$

3. Multiply the standard form equation by the I.F. ($\sin x$): $$\sin x\frac{dy}{dx}+y\cos x=x\sin x$$ The left-hand side becomes the derivative of the product $y\cdot \sin x$: $$\frac{d}{dx}(y\sin x)=x\sin x$$

4. Integrate both sides with respect to $x$: $$y\sin x=\int x\sin xdx$$

5. Evaluate the integral $\int x\sin xdx$ using Integration by Parts ($\int udv=uv-\int vdu$): Let $u=x⟹du=dx$ Let $dv=\sin xdx⟹v=-\cos x$ $$\int x\sin xdx=x(-\cos x)-\int (-\cos x)dx$$ $$=-x\cos x+\int \cos xdx$$ $$=-x\cos x+\sin x+C$$

6. Substitute back into the equation: $$y\sin x=-x\cos x+\sin x+C$$ $$y=\frac{-x\cos x+\sin x+C}{\sin x}$$ $$y=-x\cot x+1+C\csc x$$

Final Answer: The equation is a First Order Linear Differential Equation. The solution is $y=1-x\cot x+C\csc x$.

b) Evaluation of the Integral

Evaluate: $$I=\int \frac{x}{(x-1)(x^2+1)}dx$$

Step 1: Partial Fraction Decomposition Assume: $$\frac{x}{(x-1)(x^2+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}$$ Multiply by the denominator $(x-1)(x^2+1)$: $$x=A(x^2+1)+(Bx+C)(x-1)$$

Step 2: Determine Constants A, B, and C

• Put $x=1$: $$1=A(1^2+1)+0⟹1=2A⟹A=\frac{1}{2}$$
• Compare coefficients of $x^2$: $$0=A+B⟹B=-A=-\frac{1}{2}$$
• Compare constant terms: $$0=A-C⟹C=A=\frac{1}{2}$$

So, the decomposition is: $$\frac{x}{(x-1)(x^2+1)}=\frac{1/2}{x-1}+\frac{-\frac{1}{2}x+\frac{1}{2}}{x^2+1}$$ $$=\frac{1}{2}\left[\frac{1}{x-1}-\frac{x-1}{x^2+1}\right]$$ $$=\frac{1}{2}\left[\frac{1}{x-1}-\frac{x}{x^2+1}+\frac{1}{x^2+1}\right]$$

Step 3: Integrate Term by Term $$I=\frac{1}{2}\left[\int \frac{1}{x-1}dx-\int \frac{x}{x^2+1}dx+\int \frac{1}{x^2+1}dx\right]$$

1. $\int \frac{1}{x-1}dx=\ln |x-1|$
2. For $\int \frac{x}{x^2+1}dx$, let $u=x^2+1$, then $du=2xdx⟹xdx=\frac{du}{2}$. $$\int \frac{x}{x^2+1}dx=\frac{1}{2}\int \frac{1}{u}du=\frac{1}{2}\ln |x^2+1|$$
3. $\int \frac{1}{x^2+1}dx={\tan }^{-1}(x)$

Substituting these back: $$I=\frac{1}{2}\left[\ln |x-1|-\frac{1}{2}\ln (x^2+1)+{\tan }^{-1}(x)\right]+C$$ $$I=\frac{1}{2}\ln |x-1|-\frac{1}{4}\ln (x^2+1)+\frac{1}{2}{\tan }^{-1}(x)+C$$

Final Answer: $\frac{1}{2}\ln |x-1|-\frac{1}{4}\ln (x^2+1)+\frac{1}{2}{\tan }^{-1}(x)+C$

c) Rate of Change of Distance

Let the starting point be the origin $O(0,0)$. Let $x$ be the distance traveled by the car going East. Let $y$ be the distance traveled by the car going South. Let $z$ be the distance between the two cars.

Given: $$\frac{dx}{dt}=60\text{km/hr}$$ $$\frac{dy}{dt}=80\text{km/hr}$$

By the Pythagorean theorem, the relationship between the distances is: $$z^2=x^2+y^2$$

Differentiate both sides with respect to time $t$: $$2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$$ $$z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$$ $$\frac{dz}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{z}$$

Since $z=\sqrt{x^2+y^2}$, the symbolic expression for the rate at which the distance is increasing is: $$\frac{dz}{dt}=\frac{60x+80y}{\sqrt{x^2+y^2}}$$

Final Answer: The rate of increase in distance is $\frac{60x+80y}{\sqrt{x^2+y^2}}$ km/hr.