Let y=y(x) be the solution of the differential equation x3dy+(xy−1)dx=0,x>0,y(21)=3−e. Then y(1) is equal to
Your answer stays private
What feels right?
Hint 1 of 4
How is the given differential equation x3dy+(xy−1)dx=0 expressed in standard first-order linear form dxdy+P(x)y=Q(x)?
No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.
Step-by-step solutionView
Correct answer
Rewriting the differential equation in standard linear form yields the integrating factor e−1/x, giving y(x) = rac{1}{x} + 1 - e^{2 - 1/x}, which evaluates to 1 at x=1.
Option analysis
Why each option works or fails
A · 1
None. This is the correct value obtained by solving the linear differential equation with the given initial condition. Correctly identify the integrating factor I(x)=e−1/x, solve for y(x), use y(1/2)=3−e to find the constant C=−e2, and evaluate at x=1.
B · e
A student might omit the constant term or make a sign error when solving for the integration constant C, leading to an extra factor of e remaining. Carefully substitute x=1/2 into ye−1/x=(1/x+1)e−1/x+C to ensure C correctly balances with the initial condition 3−e.
C · 3
A student might calculate C=0 incorrectly by misinterpreting y(1/2)=3−e or dropping the exponential terms entirely. Keep all exponential factors intact when solving y(1/2)e−2=(2+1)e−2+C.
D · 2−e
A student might forget to evaluate the 1/x term at x=1 or subtract the terms incorrectly after evaluating the constant. Ensure that every term of the particular solution y(x) = rac{1}{x} + 1 - e^{2 - 1/x} is computed at x=1, giving 1+1−e=2−e only if C=−e rather than −e2.
Reviewed route
Solution
StepWorking
01given
The differential equation is x3dy+(xy−1)dx=0 for x>0, with initial condition y(21)=3−e.
02goal
Find the value of y(1).
03approach
Rearrange the equation into the standard linear first-order differential equation form dxdy+P(x)y=Q(x), determine the integrating factor I.F.=e∫P(x)dx, solve via substitution for the RHS integral, determine the constant C using the initial value, and finally evaluate at x=1.
04execute
Rewrite the equation: x3dxdy+xy=1⟹dxdy+x21y=x31. Here P(x)=x21 and Q(x)=x31.
The integrating factor is I.F.=e∫x21dx=e−1/x.
Multiplying gives: ye−1/x=∫x31e−1/xdx.
Substitute t=−x1, so dt=x21dx and x1=−t:
∫x31e−1/xdx=∫(−t)etdt=−(t−1)et+C=(1−t)et+C=(1+x1)e−1/x+C.
Hence, y(x)=1+x1+Ce1/x.
05execute
Use the initial condition y(21)=3−e:
3−e=1+2+Ce2=3+Ce2⟹Ce2=−e⟹C=−e−1=−e1.
Now evaluate at x=1:
y(1)=1+1+(−e1)e1=2−1=1.
✓verify
Check at x=1/2: y(1/2)=1+2−e−1⋅e2=3−e, which matches the given condition. Also, check y′(1): with y(1)=1, y′=1−1=0; from general formula y′=−1/x2−e1/x−1/x2, at x=1, y′(1)=−1−(−1)=0. Solution is completely consistent.
Hints that build this answer step by step
How is the given differential equation x3dy+(xy−1)dx=0 expressed in standard first-order linear form dxdy+P(x)y=Q(x)?
dxdy+x21y=x31
What is the integrating factor I(x)=e∫P(x)dx for this differential equation?
e−1/x
What is the general solution y⋅I(x)=∫Q(x)I(x)dx+C?
ye−1/x=(x1+1)e−1/x+C
Using y(21)=3−e, what is the value of the constant C, and what is y(1)?
Why do we make the substitution t=−1/x instead of direct integration by parts?
Direct integration by parts also works by choosing u=1/x and dv=x21e−1/xdx, but setting t=−1/x simplifies the integrand to standard polynomial times exponential ∫−tetdt, minimizing algebra slips.