Answer to Question #178600 in Differential Equations for Glad

Question

Solve the following equations using appropriate substitutions

a) y' = 5y +( e^−3x)(y^−3)

b) xdy = y(ln x − ln y)dx, y(1) = 4, x > 0.

Answer 1

Solution.

part a)

$y'-5y-e^{-3x}y^{-3}=0$

Rewrite the

original differential equation as follows:

$-5y^4e^{3x}+y^3y'e^{3x}=1$

This is the Bernoulli equation for n = 4.

Substitution $z =y^4.$

Then: $z'=4y^3y',$

and therefore the equation is rewritten as

$-5ze^{3x}+\frac{1}{4}z'e^{3x}=1.$

We solve this equation by the method of variation of an arbitrary constant.

This is a non-homogeneous equation. Consider the corresponding homogeneous equation:

$-5ze^{3x}+\frac{1}{4}z'e^{3x}=0.$

Solving it, we get:

$z'=20z.$

Integrating, we get:

$\ln{z}=20x,\newline z=e^{20x}.$

We are now looking for a solution of the original equation in the form:

$z(x)=C(x)e^{20x},\newline z '(x) = C (x) e ^{20 x} + C (x) (e ^{20 x} ) '.$

So,

$C'(x)= 4 e ^{-23 x}$

Integrating, we obtain:

$C(x)=-\frac{4}{23}e^{-23x}.$

From the condition

$z(x)=-\frac{4}{23}e^{-23x}e^{20x}.$

Since $z=y^4,$  we get:

$y^4=Ce^{20x}-\frac{4}{23}e^{-3x}.$

part b)

$$xy′=y(\ln x−\ln y).$$

We make substitution

$$u(x)=\frac{y(x)}{x}.$$

Then $y(x)=u(x)x$ and

$y'=u'x+u.$

We will have

$x^2u'+xu=ux(\ln x-\ln (ux)),$

or $x^2u'+xu\ln u+xu=0.$

Solve this equation:

$\frac{u'}{(\ln u+1)u}=-\frac{1}{x},$ or

$\frac{du}{(\ln u+1)u}=-\frac{dx}{x}.$

$\int\frac{du}{(\ln u+1)u}=-\int\frac{dx}{x},$

$\ln ({\ln u +1})=-\ln x +C,$

where C is some constant.

From here $u = e^{\frac{C_1}{x}-1}.$

And the solution of given equation is

$$y = xe^{\frac{C_1}{x}-1}.$$

y(1)=4, then

$e^{C_1-1}=4,\newline C_1-1=\ln 4,\newline C_1=\ln 4+1,\newline y=xe^{\frac{\ln 4 +1}{x}-1}.$