How do you find the particular solution to #yy'-e^x=0# that satisfies y(0)=4?

Question

Emilia Aguilar · Accepted Answer

#y(x) = sqrt(2e^x+15)#

The differential equation:

#y*(dy)/(dx) -e^x =0 #

is separable, so we can solve it by separating he variables and integrating:

#y*(dy)/(dx) = e^x #

#ydy =e^xdx#

#int ydy = int e^xdx#

#y^2/2 = e^x+C#

#y^2 = 2e^x+C#

for #x=0# we have:

#y(0)^2 = e^0+C#

#16= 1+ C => C =15#

So the particular solution we are seeking is:

#y(x) = sqrt(2e^x+15)#

William Abdalla · Answer

undefined To find the particular solution to the differential equation $yy' - e^x = 0$ that satisfies $y(0) = 4$, follow these steps:

1. Separate variables to obtain:
$$ yy' = e^x $$

2. Integrate both sides with respect to $x$:
$$ \int y \, dy = \int e^x \, dx $$

3. Integrate:
$$ \frac{1}{2} y^2 = e^x + C $$

4. Solve for $y$:
$$ y^2 = 2(e^x + C) $$

5. Use the initial condition $y(0) = 4$ to find $C$:
$$ 4^2 = 2(e^0 + C) $$
$$ 16 = 2(1 + C) $$
$$ 8 = 1 + C $$
$$ C = 7 $$

6. Substitute $C$ back into the equation:
$$ y^2 = 2(e^x + 7) $$

7. Take the square root of both sides to find $y$:
$$ y = \pm \sqrt{2(e^x + 7)} $$

Given the initial condition $y(0) = 4$, we choose the positive root:
$$ y = \sqrt{2(e^x + 7)} $$