Question

The form of the exact solution to 2dy/dx +3y= e^(-x), y(0)=5 is

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

The solution of the differential equation is  $ye^{\frac{3}{2}x}$ =  $e^{\frac{x}{2}}$  + 4

Step-by-step explanation:

The given equation is:

$2\frac{dy}{dx}$ + 3y = $e^{-x}$

Dividing the entire equation by 2

=> $\frac{dy}{dx}$ + $\frac{3y}{2}$ = $\frac{e^{-x} }{2}$

Now, this becomes an differential equation of the form:

$\frac{dy}{dx}$ + Py = Q(x)

where P =  $\frac{3}{2}$

and Q =  $\frac{e^{-x} }{2}$

We know that integrating factor (IF) = $e^{\int{P(x) dx}}$

=> IF =  $e^{\int{\frac{3}{2} dx}}$

        =  $e^{\frac{3}{2}\int{ dx}}$

        =   $e^{\frac{3}{2}x}$

Now, the equation becomes,

y * IF = $\big\int$Q * IF

=> y *  $e^{\frac{3}{2}x}$ =   $\big\int$ $\frac{e^{-x} }{2}$ * $e^{\frac{3}{2}x}$ dx

=> $ye^{\frac{3}{2}x}$ =  $\frac{1}{2}$ $\big\int$ $e^{-x}$ * $e^{\frac{3}{2}x}$ dx

=>  $ye^{\frac{3}{2}x}$ =  $\frac{1}{2}$ $\big\int$ $e^{\frac{x}{2}}$ dx

=>  $ye^{\frac{3}{2}x}$ =  $\frac{1}{2}$ * $e^{\frac{x}{2}}$ * 2 + c

=>  $ye^{\frac{3}{2}x}$ =  $e^{\frac{x}{2}}$  + c

The given initial condition is y(0) = 5

=> At x = 0, y = 5

Substituting these values in the equation above,

=>  $5*e^{\frac{3}{2}*0}$ =  $e^{\frac{0}{2}}$  + c

=> 5*1 = 1 + c

=> 5 = 1 + c

=> c = 4

Therefore, the solution of the differential equation is :  $ye^{\frac{3}{2}x}$ =  $e^{\frac{x}{2}}$  + 4