DryRun
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Homework Statement
Solve [tex]\frac{dy(t)}{dt}+2ty(t)=5t<br /> \\y(0)=1[/tex]
The attempt at a solution
Since this is a 1st order ODE, i find the integrating factor,
[tex]μ(t)=e^{\int P(t).dt}<br /> \\P(t)=2t<br /> \\Q(t)=5t<br /> \\μ(t)=e^{t^2}[/tex]
Using formula:
[tex]\frac{dμy}{dt}=μQ[/tex]
Integrating both R.H.S. and L.H.S.:
[tex]μy=\int e^{t^2}.5t\,.dt<br /> \\μy=\frac{5}{2}e^{t^2}+A<br /> \\y=\frac{5}{2}+\frac{A}{e^{t^2}}[/tex]
Using y(0)=1
When t=0, y=1
[itex]1=\frac{5}{2}+A[/itex], so A=1-(5/2)=-3/2
Therefore, the particular solution is: y=(5/2)-(3/2) which gives y=1.
Is this answer correct?
Solve [tex]\frac{dy(t)}{dt}+2ty(t)=5t<br /> \\y(0)=1[/tex]
The attempt at a solution
Since this is a 1st order ODE, i find the integrating factor,
[tex]μ(t)=e^{\int P(t).dt}<br /> \\P(t)=2t<br /> \\Q(t)=5t<br /> \\μ(t)=e^{t^2}[/tex]
Using formula:
[tex]\frac{dμy}{dt}=μQ[/tex]
Integrating both R.H.S. and L.H.S.:
[tex]μy=\int e^{t^2}.5t\,.dt<br /> \\μy=\frac{5}{2}e^{t^2}+A<br /> \\y=\frac{5}{2}+\frac{A}{e^{t^2}}[/tex]
Using y(0)=1
When t=0, y=1
[itex]1=\frac{5}{2}+A[/itex], so A=1-(5/2)=-3/2
Therefore, the particular solution is: y=(5/2)-(3/2) which gives y=1.
Is this answer correct?
Last edited: