For the given problem:` yy'-2e^x=0` , we can evaluate this by applying variable separable differential equation in which we express it in a form of `f(y) dy = f(x)dx` .
Then, `yy'-2e^x=0` can be rearrange into `yy'= 2e^x`
Express y' as (dy)/(dx):
`y(dy)/(dx)= 2e^x`
Apply direct integration in the form of `int f(y) dy = int f(x)dx` :
`y(dy)/(dx)=2e^x`
`ydy= 2e^xdx`
`int ydy= int 2e^x dx`
For the left side, we apply Power Rule integration: `int u^n du= u^(n+1)/(n+1)` .
`int y dy= y^(1+1)/(1+1)`
` = y^2/2`
For the right side, we apply basic integration property: `int c*f(x)dx= c int f(x) dx` and basic integration formula for exponential function: `int e^u du = e^u+C ` on the right side.
`int 2e^x dx= 2int e^x dx`
`= 2e^x+C`
Combining the results for the general solution of differential equation:
`y^2/2=2e^x+C`
`2* [y^2/2] = 2*[2e^x]+2*C `
Let `2*C= C` . Just a constant.
`y^2= 4e^x+C`
To find the particular solution we consider the initial condition `y(0)=3` which implies `x=0` and `y =3` .
Plug them in to `y^2= 4e^x+C` , we get:
`3^2= 4e^0+C`
`9= 4*1+C`
`9=4+C`
Then `C=9-4=5` .
Plug-in `C=5` in`y^2= 4e^x+C` , we get the particular solution:
`y^2= 4e^x+5`
`y = +-sqrt(4e^x+5).`
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