The equation of the curve passing through (0, 0) and satisfying y' = eˣ sin x is:
y(x) = ½ eˣ (sin x − cos x) + ½
We are given that the curve passes through (0, 0) and its derivative is eˣ sin x. We must find the equation of the curve.
- \frac{dy}{dx} = e^{x} \sin xchecked
We start with the differential equation for the curve.
- y(x) = \int e^{x} \sin x\,dx + Cchecked
To find y, we integrate both sides with respect to x.
- \int e^{x} \sin x\,dx = \frac{1}{2} e^{x}(\sin x - \cos x)checked
We use integration by parts or the standard result for this integral.
- y(x) = \frac{1}{2} e^{x}(\sin x - \cos x) + Cchecked
So the general solution is y(x) = ½ eˣ (sin x − cos x) + C.
- 0 = \frac{1}{2} e^{0}(\sin 0 - \cos 0) + C
We use the point (0, 0) to find the value of C.
- 0 = \frac{1}{2}(0 - 1) + C
We substitute the values: e⁰ = 1, sin 0 = 0, cos 0 = 1.
- C = \frac{1}{2}
This gives C = 1/2.
- y(x) = \frac{1}{2} e^{x}(\sin x - \cos x) + \frac{1}{2}checked
So the required equation of the curve is y(x) = ½ eˣ (sin x − cos x) + ½.
The equation of the curve is y equals one half e to the x times open bracket sin x minus cos x close bracket plus one half.
Every ticked line was worked again and checked by the computer (SymPy 1.11.1).
- Step 1: The question gives the differential equation y' = eˣ sin x. This means the slope (rate of change) of y at any x is eˣ sin x.
- Step 2: To find y itself, we need to undo the derivative. This is called integration.
- Step 3: Integrating eˣ sin x is tricky, but there is a standard result: ∫eˣ sin x dx = ½ eˣ (sin x − cos x).
- Step 4: So, the general solution is y(x) = ½ eˣ (sin x − cos x) + C, where C is a constant.
- Step 5: The curve passes through (0, 0). This means when x = 0, y = 0. We use this to find C.
- Step 6: Substitute x = 0: e⁰ = 1, sin 0 = 0, cos 0 = 1. So, y(0) = ½ × 1 × (0 − 1) + C = −½ + C.
- Step 7: Set y(0) = 0: 0 = −½ + C, so C = ½.
- Step 8: Put C = ½ in the general solution. The final answer is y(x) = ½ eˣ (sin x − cos x) + ½.
- Write the differential equation.
- Integrate both sides to get y(x) + C.
- Use the given point to find C.
- Write the final answer.
- Try x = 0:
- sin 0 = 0, cos 0 = 1, e⁰ = 1
- y(0) = ½ × 1 × (0 − 1) + ½ = (−½) + ½ = 0
- So, the curve really passes through (0, 0).
- Forgetting the constant C: Always add C after integrating.
- Wrong value for C: Not using the point (0, 0) correctly to find C.
- Mistaking the integral: Not knowing or applying the standard result for ∫eˣ sin x dx.
- Sign errors: Mixing up sin x − cos x with cos x − sin x.
Find the equation of the curve passing through (0, 2) if y' = eˣ cos x.
Check yourself
What is the value of C if the curve passes through (0, 0)?
C = ½, because substituting x = 0, y = 0 gives C = ½.
Why do we add a constant C after integrating?
Because integration gives a family of solutions; C adjusts to fit the initial condition.
If y' = eˣ sin x, what is the integral of the right side?
½ eˣ (sin x − cos x) + C, using the standard result for this integral.
How do you check if your answer is correct?
Substitute the given point into your solution and see if it satisfies the equation.
Quick recap
- Integrate both sides to solve a differential equation.
- Always add a constant C after integrating.
- Use the given point to find the value of C.
- Check your answer by substituting the point back.
- Know standard integrals like ∫eˣ sin x dx.