An illustration that complements a mathematical question. The image should visually express the concept of a differential equation, specifically focusing on dy/dx = (1+lnx)y equation. It should also indicate a condition where y=1 when x=1. Avoid any written text in this mathematical backdrop, and let the visual elements depict the equation and its condition.

if dy/dx= (1+lnx)y and if y=1 when x=1, then y=

Well, this problem seems a bit logarithmic. Let's put on our thinking caps, my friend! We have the differential equation dy/dx = (1+lnx)y, and we also know that y = 1 when x = 1.

Now, let's address the elephant in the room, the logarithm. As we know, the natural logarithm function, ln(x), goes cray-cray when x is negative or zero. But fret not, my friend, because here x is positive! So we don't have to kick out the logarithm from the party.

Now, let's work our way towards the solution. We have:
dy/dx = (1+lnx)y

The first thing we can do is separate our variables, y on one side and x stuff on the other. Just like a messy breakup, we gotta keep things separate, you know?

(dy/y) = (1+lnx) dx

Now, we can integrate both sides. The integral of the left side is simply ln(y). For the right side, we have to do a little trickery. We'll use u-substitution, setting u = lnx. Then, du = (1/x) dx, and our integral becomes:

∫(1+lnx) dx = ∫(1+u) (1/x) dx

Now, we can rewrite it as:

∫(1+u) (1/x) dx = ∫(1+u) du

Integrating, we get:

ln(y) = u + (u^2)/2 + C

Now, we just need to plug in our initial condition, y = 1 when x = 1. Plugging these values in, we get:

ln(1) = 0 = 0 + (0^2)/2 + C
C = 0

Therefore, our solution is:

ln(y) = u + (u^2)/2

Now, we just need to replace u with lnx:

ln(y) = lnx + (lnx^2)/2

Exponentiating both sides:

y = e^(lnx) * e^((lnx^2)/2)

Simplifying this expression, we get:

y = x * √(x)

And there you have it! The solution to our differential equation is y = x * √(x). Voila!

To find the value of y, we need to solve the given differential equation.

Given: dy/dx = (1 + ln(x))y

Separating the variables, we can rewrite the equation as:

(1/y) dy = (1 + ln(x)) dx

Integrating both sides,

∫(1/y) dy = ∫(1 + ln(x)) dx

ln|y| = x + xln(x) + C1, where C1 is the constant of integration.

Next, we can rewrite the equation in exponential form:

|y| = e^(x + xln(x) + C1)

Since y = 1 when x = 1, we can plug in the values and solve for the constant C1.

|1| = e^(1 + 1ln(1) + C1)

1 = e^(1 + C1)

Taking the natural logarithm on both sides,

ln(1) = ln(e^(1 + C1))

0 = 1 + C1

C1 = -1

Now we can substitute C1 back into our equation:

|y| = e^(x + xln(x) - 1)

Since |y| can be positive or negative, we have two possible solutions:

y = e^(x + xln(x) - 1) or y = -e^(x + xln(x) - 1)

Therefore, the solutions for y are y = e^(x + xln(x) - 1) and y = -e^(x + xln(x) - 1).

To solve this differential equation, we will use the technique of separating variables. The given differential equation is:

dy/dx = (1 + ln(x))y

We can rewrite this equation as:

dy/y = (1 + ln(x))dx

Now, let's integrate both sides of the equation with respect to their respective variables to separate them:

∫(dy/y) = ∫(1 + ln(x))dx

The integral of (dy/y) is ln|y| + C1, where C1 is the constant of integration.

The integral of (1 + ln(x))dx can be split into two separate integrals:

∫1dx + ∫ln(x)dx

The integral of 1dx is x + C2, where C2 is the constant of integration.

The integral of ln(x)dx can be evaluated using integration by parts, where u = ln(x) and dv = dx. By applying integration by parts, we get:

∫ln(x)dx = xln(x) - ∫(x * 1/x)dx
= xln(x) - x + C3, where C3 is the constant of integration.

Combining all the integrals:

ln|y| + C1 = x + xln(x) - x + C3

Simplifying and rearranging the equation:

ln|y| = xln(x) + C

Now, we can solve for y by exponentiating both sides of the equation:

|y| = e^(xln(x) + C)
= e^x * e^(ln(x^C))
= C * x * e^x

Since the absolute value of y appears in the equation, we need to consider both positive and negative values for C. Let's denote C as ±Ce, where Ce represents a positive constant.

Therefore, the general solution to the differential equation is:

y = ±Cx * e^x

To find the particular solution that satisfies the initial condition y = 1 when x = 1, we substitute these values into the general solution:

1 = ±C * 1 * e^1
1 = ±C * e

Depending on the sign of C (positive or negative), we have two possible solutions for y:

If C is positive: y = Ce^x

If C is negative: y = -Ce^x

Since y = 1 when x = 1, only the positive solution is valid:

y = Ce^x

Therefore, the solution to the given differential equation, with the initial condition y = 1 when x = 1, is y = Ce^x.

dy/dx= (1+lnx)y --->

dy/y = (1 + ln(x)) dx --->

ln(y) = x ln(x) + c

Boundary condition:

y=1 when x=1 ---> c = 0

y = exp(x ln(x)) = x^x