Solve the initial value problem given by dydx=2x+3,y(1)=4\frac{dy}{dx}=\frac{2}{x}+3,y\left(1\right)=4.

y = 2 ln ⁡ ∣ x ∣ + 3 x + 1 y=2\ln\left|x\right|+3x+1

Solve the initial value problem given by d y d x = 2 x + 3 , y ( 1 ) = 4 \frac{dy}{dx}=\frac{2}{x}+3,y\left(1\right)=4 .

Here we're asked to solve the initial value problem given by dy dx is equal to two over x plus three, where y of one is equal to four. So if dy over dx is equal to two over x plus three, that means that we can find our function y of x by taking the integral of two over x plus three with respect to x. So So going ahead and finding this integral here starting with that first term two over x, this is going to give us two times the natural log of the absolute value of x and then for this second term here it's just a constant so that's going to be three x. Then we want to add on that constant of integration c there. So this is what y of x is equal to, but we're given additional information. We know that y of one is equal to four, so we can use that in order to find our constant c. So if four is equal to two times the natural log of one, just plugging one everywhere x is, plus three times one, plus c. I can go ahead and use some algebra here. Now the natural log of one is going to be a zero, so that whole term goes away. And we're just left with four equals three times one, which is three, plus c. Then if I subtract three on both sides here to get it to cancel, I'm left with a one is equal to c. So now I know what my constant is, and I can plug it back into my function y of x. So doing that here, y of x is going to be equal to two times the natural log of the absolute value of x plus three x plus one of that constant we that we just found. So this is my final answer here. Let us know if you have any questions, and I'll see you in the next one.

Solve the initial value problem given by d y d x = 2 x + 3 , y ( 1 ) = 4 \frac{dy}{dx}=\frac{2}{x}+3,y\left(1\right)=4 .

y = 2 ln ⁡ ∣ x ∣ + 3 x + 1 y=2\ln\left|x\right|+3x+1

y = 2 ln ⁡ ∣ x ∣ + 3 x + 4 y=2\ln\left|x\right|+3x+4

y = 2 ln ⁡ ∣ x ∣ + 3 x y=2\ln\left|x\right|+3x

Table of contents

Skip topic navigation

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Logarithmic Functions

Calculus

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Logarithmic Functions

Previous problemNext problem

Struggling with Calculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Solve the initial value problem given by $\frac{d y}{d x} = \frac{2}{x} + 3, y (1) = 4$ .

$y = 2 \ln ∣ x ∣ + 3 x + 4$

$y = 2 \ln ∣ x ∣ + 3 x$

$y = 2 \ln ∣ x ∣ + 3 x + 1$

$y = 3 x + 1$

Previous problemNext problem

Verified step by step guidance

Identify the type of differential equation: The given equation $ \frac{dy}{dx} = \frac{2}{x} + 3 $ is a first-order ordinary differential equation. The goal is to solve for $ y $ and satisfy the initial condition $ y(1) = 4 $.

Separate the variables or recognize the structure: This equation is linear and can be solved using the method of integration. Rewrite the equation as $ dy = \left( \frac{2}{x} + 3 \right) dx $.

Integrate both sides: Integrate $ \int \frac{2}{x} dx $ and $ \int 3 dx $ separately. For $ \int \frac{2}{x} dx $, the result is $ 2 \ln|x| $. For $ \int 3 dx $, the result is $ 3x $. Combine these results to get $ y = 2 \ln|x| + 3x + C $, where $ C $ is the constant of integration.

Apply the initial condition $ y(1) = 4 $: Substitute $ x = 1 $ and $ y = 4 $ into the equation $ y = 2 \ln|x| + 3x + C $. Since $ \ln|1| = 0 $, the equation simplifies to $ 4 = 0 + 3(1) + C $. Solve for $ C $ to find its value.

Write the final solution: Substitute the value of $ C $ back into the general solution $ y = 2 \ln|x| + 3x + C $. This gives the specific solution that satisfies the initial condition.