Step by step solution to "Solve differential equationy′+3x2y=sin⁡(x)e−x3, y(0)=1"

Name: Solve differential equationy′+3x2y=sin⁡(x)e−x3, y(0)=1
Uploaded: 2022-02-23T17:22:57+00:00
Description: Solve differential equationy′+3x2y=sin⁡(x)e−x3, y(0)=1

College
Calculus and Analysis
Differential Equations
First order differential equations

Joni Kenny

Answered question

2021-02-21

Solve differential equation

y^{'} + 3 x^{2} y = \sin (x) e^{- x^{3}}, y (0) = 1

Answer & Explanation

au4gsf

Skilled2021-02-22Added 95 answers

y^{'} + p (x) y = q (x), p (x) = 3 x^{2}, q (x) = \sin (x) e^{- x^{3}}

e^{\int p d x} = e^{\int 3 x^{2} d x}

e^{\int p d x} = e^{\frac{3 x^{3}}{3}}

e^{\int p d x} = e^{x^{3}}

That is the I.F. is

μ (x) = e^{x^{3}}

Multiply the integrating factor on both sides of the given equation

e^{x^{3}} y^{'} = \sin (x) e^{- x^{3}} e^{x^{3}}

e^{x^{3}} y^{'} = \sin (x) [∵ e^{- x^{3}} e^{x^{3}} = e^{- x^{3} + x^{3}} = 1]

Integrate both sides of the above equation

\int \frac{d}{d x} (e^{x^{3}} y) d x = \int \sin (x) d x

e^{x^{3}} y = - \cos (x) + c

y = \frac{- \cos (x) + c}{e^{x^{3}}}

Now compute the value of constant C by applying the given condition as follows

y (0) = \frac{- \cos (0) + c}{e^{{(0)}^{3}}}

1 = \frac{- (1) + c}{1} [∵ \cos (0) = 1]

1 = - 1 + c

1 + 1 = c

2 = c

Now substitute the value of C for the value of y and simplify further

y = \frac{- \cos (x) + 2}{e^{x^{3}}}

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Differential Equations

Didn't find what you were looking for?