can you please help me solve this problem if dy/dx=ycosx and y=3 when x=0, then y=
dy/dx = y cos(x)
dy/y = cos(x) dx
ln y + const = sin(x) + const
ln y = sin(x) + const
ln 3 = sin(0) + const
const = ln 3
ln y = sin(x) + ln 3
ln y - ln 3 = sin(x)
ln (y/3) = sin(x)
y/3 = e^sin(x)
y = 3e^sin(x)
Sure, let's take a hilarious approach to solve this problem!
Okay, let's see. We're given that dy/dx = ycos(x), meaning the derivative of y with respect to x is equal to y times cos(x).
Now, using my clown brain, I can solve this step by step. We'll start by separating the variables, because apparently, variables need some alone time too.
dy/y = cos(x) dx
Now, hold on, let me get my giant clown calculator to perform some magical calculations.
Integrating both sides, we get:
ln|y| = sin(x) + C
Now, we have this constant C hanging around, just like a balloon at a kid's party. But don't worry, we'll handle it.
Exponentiating both sides (poof, magic!):
|y| = e^(sin(x) + C)
But if we remove those absolute value signs from |y|, we might make y cry. So let's consider two cases:
Case 1: y > 0
y = e^(sin(x) + C)
Case 2: y < 0
y = -e^(sin(x) + C)
Finally, we're given that when x = 0, y = 3. So let's plug that in:
Case 1:
3 = e^(sin(0) + C)
3 = e^(0 + C)
3 = e^C
Case 2:
3 = -e^(sin(0) + C)
3 = -e^(0 + C)
3 = -e^C
Now, let me get my magnifying clown glasses to check if there's a solution for C... *squints*
And there it is! The solution is C = ln(3) for both cases!
So, putting that C value back into the equations:
Case 1:
y = e^(sin(x) + ln(3))
y = e^(sin(x)) * 3
Case 2:
y = -e^(sin(x) + ln(3))
y = -e^(sin(x)) * 3
Voila! We have found our clown-tastic solutions. Depending on whether you like positive or negative numbers, y can be expressed as either e^(sin(x)) * 3 or -e^(sin(x)) * 3. Enjoy your math with a side of laughter!
To solve the problem, we will use separation of variables method to solve the differential equation.
Given: dy/dx = y * cos(x)
Step 1: Separate the variables by moving dx to one side and dy/y to the other side.
(dy/y) = cos(x) * dx
Step 2: Integrate both sides of the equation.
∫ (dy/y) = ∫ cos(x) * dx
Step 3: Evaluate the integrals:
ln|y| = sin(x) + C
where C is the constant of integration.
Step 4: Solve for y by exponentiating both sides of the equation:
e^(ln|y|) = e^(sin(x) + C)
|y| = e^(sin(x) + C)
Since |y| represents the absolute value of y, we have two possibilities:
- If y > 0, then y = e^(sin(x) + C)
- If y < 0, then y = -e^(sin(x) + C)
Step 5: Apply the initial condition y = 3 when x = 0 to determine the constant of integration C.
When x = 0, we have y = 3.
Substituting these values into the equation, we get:
3 = e^(sin(0) + C)
3 = e^(0 + C)
3 = e^C
Taking the natural logarithm of both sides:
ln(3) = ln(e^C)
ln(3) = C
So, the constant of integration C is ln(3).
Step 6: Substitute the value of C into the equation obtained in step 4:
- If y > 0, then y = e^(sin(x) + ln(3))
- If y < 0, then y = -e^(sin(x) + ln(3))
Step 7: Simplifying the results from step 6:
- If y > 0, then y = 3 * e^sin(x)
- If y < 0, then y = -3 * e^sin(x)
Therefore, two possible solutions for y are y = 3 * e^sin(x) or y = -3 * e^sin(x).
To solve the given differential equation dy/dx = ycos(x), we need to find the function y(x) that satisfies this equation.
We can use the method of separation of variables to solve this kind of differential equation. Here's how:
Step 1: Separate the variables.
dy/y = cos(x)dx
Step 2: Integrate both sides.
∫(1/y)dy = ∫cos(x)dx
Step 3: Evaluate the integrals.
ln|y| = sin(x) + C1
Step 4: Solve for y.
Taking the exponential of both sides, we get:
|y| = e^(sin(x) + C1)
Now, let's consider the initial condition y = 3 when x = 0.
Substituting these values into the equation, we have:
|3| = e^(sin(0) + C1)
3 = e^(0 + C1)
3 = e^C1
To determine the value of C1, we take the natural logarithm of both sides:
ln(3) = C1
Now, we can go back to the equation |y| = e^(sin(x) + C1) and substitute the value of C1:
|y| = e^(sin(x) + ln(3))
Since we're only interested in the positive solution, we can drop the absolute value signs:
y = e^(sin(x) + ln(3))
Therefore, the solution to the given differential equation with the initial condition y = 3 when x = 0 is:
y = e^(sin(x) + ln(3))