Answer :
Answer:
[tex]y=(\frac{t-17}{t-3})^{\frac{1}{14}}[/tex]
Step-by-step explanation:
We are given that initial value problem
[tex]t^2-20t+51)\frac{dy}{dt}=y[/tex]
[tex]\frac{dy}{y}=\frac{dt}{t^2-20t+51}[/tex]
[tex]\frac{dy}{y}=\frac{dt}{t^2-3t-17t+51}[/tex]
[tex]\frac{dy}{y}=\frac{dt}{(t(t-3)-17(t-3)}[/tex]
[tex]\frac{dy}{y}=\frac{dt}{(t-3)(t-17)}[/tex]
[tex]\frac{1}{(t-3)(t-17)}=\frac{A}{t-3}+\frac{B}{t-17}[/tex]
[tex]\frac{1}{(t-3)(t-17)}=\frac{A(t-17)+B(t-3)}{(t-3)(t-17)}[/tex]
[tex]1=A(t-17)+B(t-3)[/tex]...(1)
Substitute t-3=0
t=3
t-17=0
t=17
Substitute t=3 in equation (1)
[tex]1=A(3-17)+0[/tex]
[tex]1=-14A[/tex]
[tex]A=-\frac{1}{14}[/tex]
Substitute t=17
[tex]1=B(17-3)[/tex]
[tex]1=14B[/tex]
[tex]B=\frac{1}{14}[/tex]
Substitute the values of A and B
[tex]\frac{1}{(t-3)(t-17)}=-\frac{1}{14}(\frac{1}{t-3})+\frac{1}{14}(\frac{1}{t-17})[/tex]
[tex]\int\frac{dy}{y}=-\frac{1}{14}\int\frac{dt}{t-3}+\frac{1}{14}\int\frac{dt}{t-17}[/tex]
[tex]ln y=-\frac{1}{14}ln\mid{t-3}\mid+\frac{1}{14}\mid{t-17}\mid+ln C[/tex]
By using formula:[tex]\frac{dx}{x}=ln x+C[/tex]
[tex]ln y=\frac{1}{14}(-ln\mid{t-3}\mid+ln\mid{t-17}\mid)+ln C[/tex]
Using formula:[tex]ln x-ln y=ln \frac{x}{y}[/tex]
[tex]ln y=\frac{1}{14}(ln\mid{\frac{t-17}{t-3}}\mid)+ln C[/tex]
[tex]ln y=\frac{1}{14}ln\mid{\frac{t-17}{t-3}}\mid+ ln C[/tex]
Substitute y(10)=1
[tex]ln 1=\frac{1}{14}ln\mid\frac{10-17}{10-3}\mid+ln C[/tex]
[tex]0=0+ln C[/tex]
Because ln 1=0
[tex]lnC=0[/tex]
[tex]C=e^0=1[/tex]
Because [tex]ln x=y\implies x=e^y[/tex]
Substitute the value of C
[tex]ln y=\frac{1}{14}ln\mid\frac{t-17}{t-3}\mid+ln1[/tex]
[tex]ln y=\frac{1}{14}ln\mid\frac{t-17}{t-3}\mid+0[/tex]
[tex]ln y=\frac{1}{14}ln\mid\frac{t-17}{t-3}\mid[/tex]
[tex]14ln y=ln\mid\frac{t-17}{t-3}\mid[/tex]
[tex]lny^{14}=ln\mid\frac{t-17}{t-3}\mid[/tex]
By using identity [tex]blog a= loga^b[/tex]
[tex]y^{14}=\frac{t-17}{t-3}[/tex]
[tex]y=(\frac{t-17}{t-3})^{\frac{1}{14}}[/tex]
The function y as a function of t is [tex]\rm y=(\dfrac{t-17}{t-3})^{\frac{1}{14}}[/tex]
Given that
The differential equation is;
[tex]\rm (t^2-20t+51)\dfrac{dy}{dt}=y[/tex]
With y(10)= 1 y(10) =1.
We have to find
The function y is a function of t.
According to the question
The differential equation is;
[tex]\rm (t^2-20t+51)\dfrac{dy}{dt}=y[/tex]
Solving the given differential equation
[tex]\rm (t^2-20t+51)\dfrac{dy}{dt}=y\\ \\ \dfrac{dy}{y} =\dfrac{dt}{ (t^2-20t+51)}\\ \\ \dfrac{dy}{y} =\dfrac{dt}{ (t^2-3t-17t+51)}\\\\ \dfrac{dy}{y} =\dfrac{dt}{ t(t-3)-17(t-3)}\\\\ \dfrac{dy}{y} =\dfrac{dt}{ (t-3)(t-17)}\\\\ \dfrac{1}{ (t-3)(t-17)} = \dfrac{A}{t-3} + \dfrac{B}{t-17}\\ \\ \dfrac{1}{ (t-3)(t-17)} = \dfrac{A(t-17)+B(t-3)}{ (t-3)(t-17)}\\ \\ 1 = A(t-17)+B(t-3) \\ [/tex]
Here, t-3 = 0, t =3
And t-17=0, t = 17
Substitute t =3 in the equation;
[tex]\rm 1 = A(t-17)+B(t-3)\\\\ 1 = A(3-17)+B(3-3)\\ \\ 1 = -14A+0\\ \\ A = \dfrac{-1}{14}[/tex]
Substitute t =17 in the equation;
[tex]\rm 1 = A(t-17)+B(t-3)\\\\ 1 = A(17-17)+B(17-3)\\\\1 = 0+14B\\\\B= \dfrac{1}{14}[/tex]
Substitute the values of A and B in the equation;
[tex]\rm \dfrac{1}{ (t-3)(t-17)} = \dfrac{A}{t-3} + \dfrac{B}{t-17}\\\\ \\ \dfrac{1}{ (t-3)(t-17)} ={\dfrac{-1}{14}} \left(\dfrac{1}{t-3} \right) + \dfrac{1}{14} \left (\dfrac{B}{t-17} \right )\\\\\\ \dfrac{dy}{y} ={\dfrac{-1}{14}} \left(\dfrac{1}{t-3} \right) + \dfrac{1}{14} \left (\dfrac{1}{t-17} \right )[/tex]
Integrating the equation both the sides;
[tex]\rm \int\dfrac{dy}{y} ={\dfrac{-1}{14}} \int \left(\dfrac{dt}{t-3} \right) + \dfrac{1}{14} \int \left (\dfrac{dt}{t-17} \right )\\\\ lny = \dfrac{-1}{14} }ln\mid{t-3}\mid+\frac{1}{14}\mid{t-17}\mid+ln C[/tex]
By using the formula;
[tex]\rm ln x-ln y=ln \frac{x}{y}\\\\ \rm ln y=\dfrac{1}{14}(ln\mid{\frac{t-17}{t-3}}\mid)+ln C\\\\ ln y=\dfrac{1}{14}ln\mid{\frac{t-17}{t-3}}\mid+ ln C\\\\ \rm Substitute \ y(10)=1\\\\ ln 1=\dfrac{1}{14}ln\mid\frac{10-17}{10-3}\mid+ln C\\\\ 0=0+ln C\\\\Because\ ln 1=0\\\\lnC=0\\\\ C=e^0=1[/tex]
Now substitute the value of c in the equation;
[tex]\rm ln y=\frac{1}{14}ln\mid\frac{t-17}{t-3}\mid+ln1 ln y=\frac{1}{14}ln\mid\frac{t-17}{t-3}\mid+0\\\\ ln y=\frac{1}{14}ln\mid\frac{t-17}{t-3}\mid\\\\ 14ln y=ln\mid\frac{t-17}{t-3}\mid\\\\ lny^{14}=ln\mid\frac{t-17}{t-3}\mid\\\\[/tex]
By using log identity y as a function of t is;
[tex]\rm blog a= loga^b\\\\\ y^{14}=\dfrac{t-17}{t-3}\\\\ y=(\dfrac{t-17}{t-3})^{\frac{1}{14}}[/tex]
Hence, the function y as a function of t is [tex]\rm y=(\dfrac{t-17}{t-3})^{\frac{1}{14}}[/tex]
To know more about Differential Equation click the link given below.
https://brainly.com/question/14802306