Find the solution of the differential equation #xy'=y+x^2\sinx# that satisfies the given initial conditions below?

Question

#y(\pi)=0#

I know it's in #dy/dx+P(x)y=Q(x)# form, but what exactly is the integrating factor?

Like what is "exp" in #I=exp(\int P(x)dx)#?
(People use this a lot when mentioning the integrating factor, but I learned it as #I(x)=e^(\intP(x)dx)#...)

Aurora Alvarado · Accepted Answer

Refer to the Explanation.

#exp# in #exp(intP(x)dx)# means exponential function. So, #exp(intP(x)dx)# is just another way to denote #e^(intP(x)dx)#. I think I've allayed your curiosity! By the way, I'm assuming you are familiar with solving the provided diff. eqn., so I'm unable to solve it.

Charles Anctil · Answer

undefined To solve the differential equation $xy' = y + x^2 \sin(x)$ with the initial condition $y(1) = 2$, you can use the method of integrating factors. First, rewrite the equation in standard form $y' + P(x)y = Q(x)$, where $P(x)$ is the coefficient of $y$ and $Q(x)$ is the remaining terms:

$$y' - \frac{1}{x}y = x \sin(x)$$

Then, identify $P(x) = -\frac{1}{x}$ and $Q(x) = x \sin(x)$. Next, find the integrating factor $I(x)$, which is given by:

$$I(x) = e^{\int P(x) \, dx}$$

Calculate the integral:

$$\int -\frac{1}{x} \, dx = -\ln|x|$$

So, the integrating factor is $I(x) = e^{-\ln|x|} = \frac{1}{|x|}$.

Now, multiply both sides of the differential equation by the integrating factor:

$$\frac{1}{|x|} y' - \frac{1}{|x|^2}y = \sin(x)$$

This can be rewritten as:

$$\frac{d}{dx} \left(\frac{y}{|x|}\right) = \sin(x)$$

Integrate both sides with respect to $x$:

$$\frac{y}{|x|} = -\cos(x) + C$$

Where $C$ is the constant of integration.

Now, multiply both sides by $|x|$ to solve for $y$:

$$y = -|x| \cos(x) + C|x|$$

Now, apply the initial condition $y(1) = 2$ to find the value of $C$:

$$2 = -1 \cdot \cos(1) + C \cdot 1$$

$$C = 2 + \cos(1)$$

So, the solution to the differential equation with the given initial condition is:

$$y = -|x| \cos(x) + (2 + \cos(1)) |x|$$

Find the solution of the differential equation #xy'=y+x^2\sinx# that satisfies the given initial conditions below?

#y(\pi)=0# I know it's in #dy/dx+P(x)y=Q(x)# form, but what exactly is the integrating factor? Like what is "exp" in #I=exp(\int P(x)dx)#? (People use this a lot when mentioning the integrating factor, but I learned it as #I(x)=e^(\intP(x)dx)#...)

#y(\pi)=0#

I know it's in #dy/dx+P(x)y=Q(x)# form, but what exactly is the integrating factor?

Like what is "exp" in #I=exp(\int P(x)dx)#?
(People use this a lot when mentioning the integrating factor, but I learned it as #I(x)=e^(\intP(x)dx)#...)