Let y=y(x) be the solution curve of the differential equation x(x2+ex)dy+(ex(x−2)y−x3)dx=0, x>0, passing through the point (1,0). Then y(2) is equal to:
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Step-by-Step Solution
Step 1: Rearrange the differential equation into standard form
First, we rearrange the given differential equation to isolate the derivative term and group terms involving y. This helps us identify if it's a linear differential equation, which can be solved using an integrating factor.
Step 2: Convert to linear first-order form
Divide the entire equation by x(x2+ex) to bring it into the standard form of a linear first-order differential equation: dxdy+P(x)y=Q(x). Here, P(x)=x(x2+ex)ex(x−2) and Q(x)=x2+exx2.
Step 3: Calculate the integrating factor
The integrating factor (I.F.) is e∫P(x)dx. We need to evaluate the integral of P(x). Notice that the numerator ex(x−2)=xex−2ex is the derivative of xex−2ex. The denominator is x(x2+ex)=x3+xex. This integral is tricky, but we can observe that the derivative of the denominator x3+xex is 3x2+ex+xex. This doesn't directly simplify. Let's re-examine the numerator. The derivative of x2+ex is 2x+ex. This also doesn't directly simplify. Let's try a different approach for the integral. Consider the derivative of ln(x2+ex). It is x2+ex2x+ex. This is not matching. Let's try to rewrite P(x) as x(x2+ex)ex(x−2)=x3+xexxex−2ex. This integral is not straightforward. Let's recheck the problem statement for any simplification. The given differential equation is x(x2+ex)dy+(ex(x−2)y−x3)dx=0. This can be written as x(x2+ex)dy=−(ex(x−2)y−x3)dx. So, dxdy=x(x2+ex)x3−ex(x−2)y. This is dxdy+x(x2+ex)ex(x−2)y=x2+exx2. Let's re-evaluate the integral of P(x). We can rewrite P(x) as x(x2+ex)ex(x−2)=x3+xexxex−2ex. This is not easily integrable. Let's try to rewrite the original equation in a different form.
Let's try to rewrite the original equation as x(x2+ex)dy+(xex−2ex)ydx=x3dx.
Divide by x2+ex: xdy+x2+exxex−2exydx=x2+exx3dx. This is not a standard form.
Let's go back to the linear form: dxdy+x(x2+ex)ex(x−2)y=x2+exx2.
The integrating factor is e∫x(x2+ex)ex(x−2)dx.
Let's try to simplify the integrand: x(x2+ex)ex(x−2)=x3+xexxex−2ex.
Consider the derivative of ln(x2+ex). It is x2+ex2x+ex.
Consider the derivative of ln(x). It is x1.
Let's try to manipulate the integrand: x(x2+ex)ex(x−2)=x(x2+ex)ex(x2+ex)−ex(x2+2). This is not working.
Let's try to rewrite the original equation as x(x2+ex)dy+(xex−2ex)ydx=x3dx.
This is x(x2+ex)dy+(xex−2ex)ydx=x3dx.
Divide by x2: (x+xex)dy+(xex−x22ex)ydx=xdx.
This is not a linear form.
Let's re-examine the original equation: x(x2+ex)dy+(ex(x−2)y−x3)dx=0.
This can be written as x(x2+ex)dy+ex(x−2)ydx=x3dx.
Divide by x2: (x+xex)dy+(xex−x22ex)ydx=xdx.
This is not a linear form.
Let's try to rewrite the equation as Mdx+Ndy=0.
M=ex(x−2)y−x3 and N=x(x2+ex).
∂y∂M=ex(x−2).
∂x∂N=(x2+ex)+x(2x+ex)=x2+ex+2x2+xex=3x2+ex+xex.
Since ∂y∂M=∂x∂N, the equation is not exact.
Let's try to find an integrating factor.
Consider N1(∂y∂M−∂x∂N)=x(x2+ex)1(ex(x−2)−(3x2+ex+xex))=x(x2+ex)xex−2ex−3x2−ex−xex=x(x2+ex)−3x2−3ex=x(x2+ex)−3(x2+ex)=−x3.
This is a function of x only. So, the integrating factor is e∫−x3dx=e−3lnx=elnx−3=x−3=x31.
Multiply the original equation by x31:
x31[x(x2+ex)dy+(ex(x−2)y−x3)dx]=0 x2x2+exdy+(x3ex(x−2)y−1)dx=0 (x3ex(x−2)y−1)dx+(1+x2ex)dy=0.
Now, let M′=x3ex(x−2)y−1 and N′=1+x2ex.
∂y∂M′=x3ex(x−2).
∂x∂N′=x4x2ex−ex(2x)=x3xex−2ex=x3ex(x−2).
Since ∂y∂M′=∂x∂N′, the equation is exact.
Now we need to solve the exact differential equation.
We know that there exists a function f(x, y) such that ∂x∂f=M′ and ∂y∂f=N′.
Integrate N′ with respect to y:
f(x,y)=∫(1+x2ex)dy=y+x2exy+g(x).
Now differentiate f(x, y) with respect to x and equate it to M′:
∂x∂f=x4x2ex−ex(2x)y+g′(x)=x3xex−2exy+g′(x).
We have M′=x3ex(x−2)y−1.
So, x3ex(x−2)y+g′(x)=x3ex(x−2)y−1.
This implies g′(x)=−1.
Integrating g'(x) with respect to x: g(x)=∫−1dx=−x+C1.
So, the solution is y+x2exy−x=C.
y(1+x2ex)=x+C.
y(x2x2+ex)=x+C.
y=x2+exx2(x+C).
Step 4: Apply the initial condition
The solution curve passes through the point (1,0). Substitute x=1 and y=0 into the general solution to find the value of the constant C.
Step 5: Write the particular solution
Substitute the value of C=−1 back into the general solution to obtain the particular solution for the given initial condition.
Step 6: Calculate y(2)
Finally, substitute x=2 into the particular solution to find the value of y(2).