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SVETLANKA909090 [29]
4 years ago
15

. Solve the following initial value problem: (t2−20t+51)dydt=y (t2−20t+51)dydt=y with y(10)=1y(10)=1. (Find yy as a function of

tt.)
Mathematics
1 answer:
Semenov [28]4 years ago
6 0

Answer:

y=(\frac{t-17}{t-3})^{\frac{1}{14}}

Step-by-step explanation:

We are given that initial value problem

t^2-20t+51)\frac{dy}{dt}=y

\frac{dy}{y}=\frac{dt}{t^2-20t+51}

\frac{dy}{y}=\frac{dt}{t^2-3t-17t+51}

\frac{dy}{y}=\frac{dt}{(t(t-3)-17(t-3)}

\frac{dy}{y}=\frac{dt}{(t-3)(t-17)}

\frac{1}{(t-3)(t-17)}=\frac{A}{t-3}+\frac{B}{t-17}

\frac{1}{(t-3)(t-17)}=\frac{A(t-17)+B(t-3)}{(t-3)(t-17)}

1=A(t-17)+B(t-3)...(1)

Substitute t-3=0

t=3

t-17=0

t=17

Substitute t=3 in equation (1)

1=A(3-17)+0

1=-14A

A=-\frac{1}{14}

Substitute t=17

1=B(17-3)

1=14B

B=\frac{1}{14}

Substitute the values of A and B

\frac{1}{(t-3)(t-17)}=-\frac{1}{14}(\frac{1}{t-3})+\frac{1}{14}(\frac{1}{t-17})

\int\frac{dy}{y}=-\frac{1}{14}\int\frac{dt}{t-3}+\frac{1}{14}\int\frac{dt}{t-17}

ln y=-\frac{1}{14}ln\mid{t-3}\mid+\frac{1}{14}\mid{t-17}\mid+ln C

By using formula:\frac{dx}{x}=ln x+C

ln y=\frac{1}{14}(-ln\mid{t-3}\mid+ln\mid{t-17}\mid)+ln C

Using formula:ln x-ln y=ln \frac{x}{y}

ln y=\frac{1}{14}(ln\mid{\frac{t-17}{t-3}}\mid)+ln C

ln y=\frac{1}{14}ln\mid{\frac{t-17}{t-3}}\mid+ ln C

Substitute y(10)=1

ln 1=\frac{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

Because ln x=y\implies x=e^y

Substitute the value of C

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

By using identity blog a= loga^b

y^{14}=\frac{t-17}{t-3}

y=(\frac{t-17}{t-3})^{\frac{1}{14}}

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