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