Question

Solve the differential equations using integrating factor.

1. x dy/dx - y = x raise to power of 3 cosx, given that y=0 and x=π.

2. ( 1 + x square) dy/dx + 3xy = 5x given y=2 and x=1

Answer 1

1. In the first case we can even guess the integrating factor (as we can see the derivative of a quotient on the left). Let's calculate it directly anyway, first we will divide both sides by $x$ to find $y' + (-\frac{1}{x}) y = x^2 \cos x$ : $P(x)=-\frac{1}{x}$, $I(x) = e^{\int P(t)dt} = \frac{1}{x}$. By multiplying both sides by $I(x)$ we find $\frac{y'}{x}-\frac{y}{x^2}=x\cos x$. Now we have $(\frac{y}{x})' = x \cos x$. By integrating both sides we find $\frac{y}{x} = x\sin x+\cos x + C$ and thus $y(x) =x^2 \sin x + x\cos x + Cx$. Let's find the constant : $0 = \pi^2 \cdot 0 + \pi \cdot (-1) + C \pi$ and so $C=1$, $y(x)=x^2 \sin x + x\cos x + x$.
2. Let's start by dividing both sides by $(1+x^2)$ : $y'+\frac{3x}{1+x^2} y = \frac{5x}{1+x^2}$. Thus $P(x) = \frac{3x}{1+x^2}$, $I(x) = e^{\int P(t) dt} = e^{\frac{3}{2} \ln(1+x^2)}=(1+x^2)^{3/2}$. Now by multyplying both sides by $I(x)$ we find $y'(1+x^2)^{3/2} +3x\cdot (1+x^2)^{1/2}y=5x\cdot (1+x^2)^{1/2}$, $((1+x^2)^{3/2}y)' = 5x \cdot (1+x^2)^{1/2}$. Now by integrating both sides we find $(1+x^2)^{3/2} y = \frac{5}{2} \cdot \frac{2}{3} (1+x^2)^{3/2} + C$ and so $y(x)=\frac{5}{3} + \frac{C}{(1+x^2)^{3/2}}$, now let's find the constant : $y(1) = \frac{5}{3} + \frac{C}{2^{3/2}} = 2$, $C = \frac{2\sqrt{2}}{3}$ and thus the solution is $y(x) = \frac{5}{3} + \frac{2\sqrt{2}}{3(1+x^2)^{3/2}}$.